Pengguna:Dedhert.Jr/Uji halaman 01/14

Dalam matematika, deret harmonik adalah deret takhingga yang dibentuk dengan menjumlahkan semua pecahan satuan positif: $\sum _{n=1}^{\infty }{\frac {1}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots .$ Jumlah suku $n$ pertama dari deret tersebut kira-kira sama dengan $\ln n+\gamma$ , dimana $\ln$ adalah logaritma alami dan $\gamma \approx 0.577$ adalah konstanta Euler–Mascheroni. Karena nilai logaritma $\ln$ merupakan nilai yang besar sebarang, deret harmonik tidak memiliki batas terhingga, sehingga deret harmonik merupakan deret divergen. Kedivergenan deret harmonik dibuktikan oleh Nicole Oresme pada abad ke-14 menggunakan suku sebelumnya dengan uji kondensasi Cauchy, uji kekonvergenan deret takhingga. Deret ini juga dapat dibuktikan berupa deret divergen dengan uji integral kekonvergenan, uji yang membandingkan penjumlahan dengan integral.

Deret harmonik beserta jumlah parsialnya dapat diterapkan pada bukti Euler bahwa ada tak berhingga banyaknya bilangan prima, analisis masalah pengumpul kupon yang menanyakan berapa banyak percobaan acak yang diperlukan agar to provide a complete range of responses, komponen terhubung graf acak, masalah penumpukan balok yang menanyakan seberapa jauh penumpukan balok di tepi meja dapat menopang, dan analisis kasus rata-rata algoritma quicksort.

Sejarah

Sebuah gelombang dan harmoniknya, dengan panjang gelombang

1,{\tfrac {1}{2}},{\tfrac {1}{3}},\dots

Nama deret harmonik berasal dari konsep nada tambahan atau harmonik dalam musik: panjang gelombang nada tambahan dari sebuah dawai yang bergetar adalah ${\tfrac {1}{2}}$ , ${\tfrac {1}{3}}$ , ${\tfrac {1}{4}}$ , dst., dari panjang gelombang dasar dawai.^[1]^[2] Setiap suku deret harmonik setelah suku pertama disebut purata harmonik suku-suku tetangga; istilah purata harmonik dan barisan harmonik juga berasal dari musik.^[2] Selain musik, barisan harmonik juga mempunyai kepopuleran dalam arsitek. Barisan harmonik khususnya dipakai pada masa periode Baroque, ketika para arsitek memakainnya untuk menggambar perbandingan denah lantai dan elevasi, serta menggambar kaitan harmonik antara detail arsitektural gereja dan kastil di bagian dalam dan di bagian luar.^{[terj. masih kasar, butuh diperhalus]}^[3]

Divergensi dari deret harmonik dibuktikan pertama kali oleh Nicole Oresme pada tahun 1350.^[2]^[4] Karya Oresme dan karya teman sebayanya, Richard Swineshead yang mengerjakan bukti tersebut pada deret yang berbeda, mengatakan kemunculan deret takhingga untuk pertama kalinya selain deret geometri dalam matematika.^[5] However, this achievement fell into obscurity.^[6] Bukti tambahannya diterbitkan abad ke-17 oleh Pietro Mengoli^[2]^[7] dan oleh Jacob Bernoulli.^[8]^[9]^[10] Bernoulli credited his brother Johann Bernoulli for finding the proof,^[10] and it was later included in Johann Bernoulli's collected works.^[11]

Jumlah parsial dari deret harmonik dinamai bilangan harmonik. Bilangan harmonik biasanya diberi notasi $H_{n}$ oleh Donald Knuth pada tahun 1968.^[12]

Definisi dan kedivergenan deret harmonik

Deret harmonik merupakan deret takhingga yang ditulis sebagai $\sum _{n=1}^{\infty }{\frac {1}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots$ Deret ini merupakan deret yang semua sukunya adalah pecahan satuan positif. Deret ini merupakan deret divergen karena semakin banyak suku deret tersebut yang memuat dalam jumlah parsial pada deret, nilai dari jumlah parsial tersebut semakin membesar sebarang, di luar suatu limit terhingga. Karena deret harmonik divergen, deretnya dapat dipandang sebagai jumlah formal, sebuah bentuk matematika yang abstrak dengan menggabungkan pecahan satuan, daripada sebagai sesuatu yang dapat dihitung sebagai nilai numerik. Ada banyak bukti yang berbeda mengenai kedivergenan deret harmonik, yang diselidiki dalam makalah S. J. Kifowit dan T. A. Stamps tahun 2006.^[13] Berikut adalah dua bukti mengenai kedivergenan deret harmonik yang terkenal.^[1]^[13]

Uji perbandingan

Cara untuk membuktikan kedivergenan deret harmonik adalah membandingkan deret harmonik dengan deret divergen lainnya, yang setiap penyebut diganti dengan perpangkatan bilangan dua terbesar selanjutnya: ${\begin{alignedat}{8}1&+{\frac {1}{2}}&&+{\frac {1}{3}}&&+{\frac {1}{4}}&&+{\frac {1}{5}}&&+{\frac {1}{6}}&&+{\frac {1}{7}}&&+{\frac {1}{8}}&&+{\frac {1}{9}}&&+\cdots \\[5pt]{}\geq 1&+{\frac {1}{2}}&&+{\frac {1}{\color {red}{\mathbf {4} }}}&&+{\frac {1}{4}}&&+{\frac {1}{\color {red}{\mathbf {8} }}}&&+{\frac {1}{\color {red}{\mathbf {8} }}}&&+{\frac {1}{\color {red}{\mathbf {8} }}}&&+{\frac {1}{8}}&&+{\frac {1}{\color {red}{\mathbf {16} }}}&&+\cdots \\[5pt]\end{alignedat}}$ Mengelompokkan suku-suku yang sama pada deret kedua memperlihatkan bahwa deret tersebut divergen: ${\begin{aligned}&1+\left({\frac {1}{2}}\right)+\left({\frac {1}{4}}+{\frac {1}{4}}\right)+\left({\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}\right)+\left({\frac {1}{16}}+\cdots +{\frac {1}{16}}\right)+\cdots \\[5pt]{}={}&1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+\cdots .\end{aligned}}$ Karena setiap suku deret harmonik lebih besar sama dengan suku yang berpadanan dengan deret kedua, maka (menurut uji perbandingan) deret harmonik juga divergen. Argumen yang serupa memberikan bukti yang lebih kuat lagi, yang mengatakan bahwa untuk setiap bilangan bulat positif $k$ , $\sum _{n=1}^{2^{k}}{\frac {1}{n}}\geq 1+{\frac {k}{2}}$ Argumen ini merupakan bukti asli Nicole Oresme sekitar tahun 1350.^[13] Uji kondensasi Cauchy merupakan perumuman dari argumen di atas.^[14]

Uji integral

Rectangles with area given by the harmonic series, and the hyperbola

y=1/x

through the upper left corners of these rectangles

Uji ini dapat membuktikan bahwa deret harmonik adalah divergen dengan membandingkan penjumlahannya dengan integral takwajar. Khususnya, uji ini dapat membuktikan dengan meninjau susunan persegi panjang seperti gambar di samping. Setiap persegi panjang mempunyai ukuran lebar 1 satuan dan tinggi ${\tfrac {1}{n}}$ satuan. Jika deret takhingga konvergen, maka total luas persegi merupakan jumlah dari deret harmonik. Kurva $y={\tfrac {1}{x}}$ sepenuhnya tetap berada di bawah batas atas persegi panjang, sehingga luas di bawah kurva (dengan domain $x$ yang berkisar dari 1 ke takhingga yang ditutupi^[?] oleh persegi panjang) lebih kecil dari luas gabungan persegi panjang. Namun, luas di bawah kurva dapat dhitung melalui integral takwajar divergen, $\int _{1}^{\infty }{\frac {1}{x}}\,dx=\infty .$ Karena integral tersebut tidak konvergen, maka jumlah dari luas persegi panjang juga tidak konvergen.^[13]

Menggantikan setiap persegi panjang selanjutnya pada barisan tersebut akan memberikan sebuah barisan persegi panjang yang panjangnya terletak di bawah kurva, bukan di atas kurva. Ini memperlihatkan bahwa jumlah parsial dari deret harmonik berbeda dengan integral by an amount that is bounded above and below by the unit area of the first rectangle: $\int _{1}^{N+1}{\frac {1}{x}}\,dx<\sum _{i=1}^{N}{\frac {1}{i}}<\int _{1}^{N+1}{\frac {1}{x}}\,dx+1.$ Generalizing this argument, any infinite sum of values of a monotone decreasing positive function of $n$ (like the harmonic series) has partial sums that are within a bounded distance of the values of the corresponding integrals. Therefore, the sum converges if and only if the integral over the same range of the same function converges. When this equivalence is used to check the convergence of a sum by replacing it with an easier integral, it is known as the integral test for convergence.^[15]

Jumlah parsial

Menambahkan suku $n$ pertama dari deret harmonik memberikan jumlah parsial, yang disebut sebagai bilangan harmonik. Bilangan harmonik biasanya dilambangkan $H_{n}$ :^[12] $H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}.$

Kelajuan pertumbuhan

Melalui pertumbuhan logaritmik, bilangan harmonik tumbuh dengan sangat lambat seperti yang diperlihatkan pada uji integral di atas.^[15] Lebih tepatnya, $H_{n}=\ln n+\gamma +{\frac {1}{2n}}-\varepsilon _{n}$ dimana $\gamma \approx 0.5772$ adalah konstanta Euler–Mascheroni dan $0\leq \varepsilon _{n}\leq 1/8n^{2}$ . Nilai $\varepsilon _{n}$ mendekati 0 ketika $n$ menuju takhingga.^[16]

Keterbagian

Bilangan harmonik bukan merupakan bilangan bulat untuk $n>1$ , terkecuali $H_{1}=1$ .^[17]^[18] Salah satu cara untuk membuktikan bahwa $H_{n}$ bukanlah bilangan bulat adalah dengan meninjau perpangkatan bilangan dua $2^{k}$ tertinggi yang berkisar antara 1 ke $n$ . Jika $M$ adalah kelipatan persekutuan terkecil dari bilangan antara 1 ke $n$ , maka $H_{k}$ dapat ditulis ulang sebagai jumlah pecahan yang memiliki penyebut yang sama $H_{n}=\sum _{i=1}^{n}{\tfrac {M/i}{M}}$ Salah satu pembilang pada pecahan tersebut hanya $M/2^{k}$ , yang berupa bilangan ganjil^[^{masih perlu diperbaiki lagi]} dan sisanya adalah bilangan genap, dan (ketika $n>1$ ) $M$ sendiri adalah bilangan genap. Oleh karena itu, hasilnya berupa sebuah pecahan dengan pembilang ganjil dan penyebut genap, yang berarti pecahan bukan bilangan bulat.^[17] Lebih lanjut, suatu barisan bilangan bulat berurutan mempunyai anggota tunggal yang dapat dibagi oleh perpangkatan bilangan dua yang lebih beasr dari semua anggota barisan lainnya, from which it follows by the same argument that no two harmonic numbers differ by an integer.^[18]

Bukti lain bahwa bilangan harmonik bukan bilangan bulat mengamati bahwa pembilang dari $H_{n}$ harus dapat dibagi oleh semua bilangan prima yang lebih besar dari $n/2$ , dan memggunakan postulat Bertrand untuk membuktikan bahwa himpunan bilangan prima adalah takkosong. Argumen yang sama menyiratkan lebih lanjut bahwa bilangan harmonik tidak dapat mewakili desimal berhenti, kecuali $H_{1}=1$ , $H_{2}=1.5$ , dan $H_{6}=2.45$ .^[17] It has been conjectured that every prime number divides the numerators of only a finite subset of the harmonic numbers, but this remains unproven.^[19]

Interpolasi

Fungsi digamma di bilangan kompleks

Fungsi digamma didefinisikan sebagai turunan logaritmik dari fungsi gamma $\psi (x)={\frac {d}{dx}}\ln {\big (}\Gamma (x){\big )}={\frac {\Gamma '(x)}{\Gamma (x)}}.$ Sama seperti fungsi gamma yang menyediakan^{[kurang tepat, ada padanan lain?]} interpolasi kontinu faktorial, fungsi digamma menyediakan^{[kurang tepat, ada padanan lain?]} interpolasi kontinu bilangan harmonik. Dengan kata lain bahwa $\psi (n)=H_{n-1}-\gamma$ .^[20]

Penerapan

Ada banyak masalah-masalah matematika yang terkenal mempunyai solusi yang melibatkan deret harmonik beserta jumlah parsialnya.

Melintasi gurun

Masalah jip atau masalah melintasi gurun merupakan kumpulan masalah pada abad ke-19 yang diatribusi oleh Alcuin dalam manuskrip Propositiones ad Acuendos Juvenes, (diformulasikan sebagai unta daripada jip), namun solusi yang diberikan salah.^[21] Masalah ini menanyakan how far into the desert a jeep can travel and return, starting from a base with $n$ loads of fuel, by carrying some of the fuel into the desert and leaving it in depots. The optimal solution involves placing depots spaced at distances ${\tfrac {r}{2n}},{\tfrac {r}{2(n-1)}},{\tfrac {r}{2(n-2)}},\dots$ from the starting point and each other, where $r$ is the range of distance that the jeep can travel with a single load of fuel. On each trip out and back from the base, the jeep places one more depot, refueling at the other depots along the way, and placing as much fuel as it can in the newly placed depot while still leaving enough for itself to return to the previous depots and the base. Therefore, the total distance reached on the $n$ th trip is ${\frac {r}{2n}}+{\frac {r}{2(n-1)}}+{\frac {r}{2(n-2)}}+\cdots ={\frac {r}{2}}H_{n},$ dengan $H_{n}$ adalah bilangan harmonik ke- $n$ . Divergensi deret harmonik menyiratikan The divergence of the harmonic series implies that crossings of any length are possible with enough fuel.^[22]

For instance, for Alcuin's version of the problem, $r=30$ : a camel can carry 30 measures of grain and can travel one leuca while eating a single measure, where a leuca is a unit of distance roughly equal to 23 kilometer (14 mi). The problem has $n=3$ : there are 90 measures of grain, enough to supply three trips. For the standard formulation of the desert-crossing problem, it would be possible for the camel to travel ${\tfrac {30}{2}}{\bigl (}{\tfrac {1}{3}}+{\tfrac {1}{2}}+{\tfrac {1}{1}})=27.5$ leucas and return, by placing a grain storage depot 5 leucas from the base on the first trip and 12.5 leucas from the base on the second trip. However, Alcuin instead asks a slightly different question, how much grain can be transported a distance of 30 leucas without a final return trip, and either strands some camels in the desert or fails to account for the amount of grain consumed by a camel on its return trips.^[21]

Masalah penumpukan balok

The block-stacking problem: blocks aligned according to the harmonic series can overhang the edge of a table by the harmonic numbers

In the block-stacking problem, one must place a pile of $n$ identical rectangular blocks, one per layer, so that they hang as far as possible over the edge of a table without falling. The top block can be placed with ${\tfrac {1}{2}}$ of its length extending beyond the next lower block. If it is placed in this way, the next block down needs to be placed with at most ${\tfrac {1}{2}}\cdot {\tfrac {1}{2}}$ of its length extending beyond the next lower block, so that the center of mass of the top two block is supported and they do not topple. The third block needs to be placed with at most ${\tfrac {1}{2}}\cdot {\tfrac {1}{3}}$ of its length extending beyond the next lower block, and so on. In this way, it is possible to place the $n$ blocks in such a way that they extend ${\tfrac {1}{2}}H_{n}$ lengths beyond the table, where $H_{n}$ is the $n$ th harmonic number.^[23]^[24] The divergence of the harmonic series implies that there is no limit on how far beyond the table the block stack can extend.^[24] For stacks with one block per layer, no better solution is possible, but significantly more overhang can be achieved using stacks with more than one block per layer.^[25]

Menghitung bilangan prima beserta keterbagiannya

Pada tahun 1737, Leonhard Euler mengamati bahwa deret harmonik sebagai jumlah formal sama dengan darab Euler. Setiap suku-suku pada darab berasal dari bilangan prima: $\sum _{i=1}^{\infty }{\frac {1}{i}}=\prod _{p\in \mathbb {P} }\left(1+{\frac {1}{p}}+{\frac {1}{p^{2}}}+\cdots \right)=\prod _{p\in \mathbb {P} }{\frac {1}{1-1/p}},$ dimana $\mathbb {P}$ melambangkan himpunan bilangan prima. Persamaan sebelah kiri dapat menerapkan hukum distributif to the product and recognizing the resulting terms as the prime factorizations of the terms in the harmonic series, dan persamaan sebelah kanan menggunakan rumus deret geometri biasa. Darabnya divergen, begitu pula dengan penjumlahannya. Namun jika darabnya konvergen, maka persamaan dapat mengambil logaritma alami dan memperoleh $\ln \prod _{p\in \mathbb {P} }{\frac {1}{1-1/p}}=\sum _{p\in \mathbb {P} }\ln {\frac {1}{1-1/p}}=\sum _{p\in \mathbb {P} }\left({\frac {1}{p}}+{\frac {1}{2p^{2}}}+{\frac {1}{3p^{3}}}+\cdots \right)=\sum _{p\in \mathbb {P} }{\frac {1}{p}}+K.$ Pada persamaan di atas, setiap logaritma digantikan dengan deret Taylor, dan konstanta $K$ pada persamaan sebelah kanan merupakan the evaluation of the convergent series of terms with exponent greater than one. It follows from these manipulations that the sum of reciprocals of primes, on the right hand of this equality, must diverge, for if it converged these steps could be reversed to show that the harmonic series also converges, which it does not. An immediate corollary is that there are infinitely many prime numbers, because a finite sum cannot diverge.^[26] Although Euler's work is not considered adequately rigorous by the standards of modern mathematics, it can be made rigorous by taking more care with limits and error bounds.^[27] Euler's conclusion that the partial sums of reciprocals of primes grow as a double logarithm of the number of terms has been confirmed by later mathematicians as one of Mertens' theorems,^[28] and can be seen as a precursor to the prime number theorem.^[27]

Another problem in number theory closely related to the harmonic series concerns the average number of divisors of the numbers in a range from 1 to $n$ , formalized as the average order of the divisor function, ${\frac {1}{n}}\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor \leq {\frac {1}{n}}\sum _{i=1}^{n}{\frac {n}{i}}=H_{n}.$ The operation of rounding each term in the harmonic series to the next smaller integer multiple of ${\tfrac {1}{n}}$ causes this average to differ from the harmonic numbers by a small constant, and Peter Gustav Lejeune Dirichlet showed more precisely that the average number of divisors is $\ln n+2\gamma -1+O(1/{\sqrt {n}})$ (expressed in big O notation). Bounding the final error term more precisely remains an open problem, known as Dirichlet's divisor problem.^[29]

Masalah pengumpulan kupon

Graph of number of items versus the expected number of trials needed to collect all items

Several common games or recreations involve repeating a random selection from a set of items until all possible choices have been selected; these include the collection of trading cards^[30]^[31] and the completion of parkrun bingo, in which the goal is to obtain all 60 possible numbers of seconds in the times from a sequence of running events.^[32] More serious applications of this problem include sampling all variations of a manufactured product for its quality control,^[33] and the connectivity of random graphs.^[34] In situations of this form, once there are $k$ items remaining to be collected out of a total of $n$ equally-likely items, the probability of collecting a new item in a single random choice is $k/n$ and the expected number of random choices needed until a new item is collected is $n/k$ . Summing over all values of $k$ from $n$ down to 1 shows that the total expected number of random choices needed to collect all items is $nH_{n}$ , where $H_{n}$ is the $n$ th harmonic number.^[35]

Menganalisis algoritma

Animation of the average-case version of quicksort, with recursive subproblems indicated by shaded arrows and with pivots (red items and blue lines) chosen as the last item in each subproblem

The quicksort algorithm for sorting a set of items can be analyzed using the harmonic numbers. The algorithm operates by choosing one item as a "pivot", comparing it to all the others, and recursively sorting the two subsets of items whose comparison places them before the pivot and after the pivot. In either its average-case complexity (with the assumption that all input permutations are equally likely) or in its expected time analysis of worst-case inputs with a random choice of pivot, all of the items are equally likely to be chosen as the pivot. For such cases, one can compute the probability that two items are ever compared with each other, throughout the recursion, as a function of the number of other items that separate them in the final sorted order. If items $x$ and $y$ are separated by $k$ other items, then the algorithm will make a comparison between $x$ and $y$ only when, as the recursion progresses, it picks $x$ or $y$ as a pivot before picking any of the other $k$ items between them. Because each of these $k+2$ items is equally likely to be chosen first, this happens with probability ${\tfrac {2}{k+2}}$ . The total expected number of comparisons, which controls the total running time of the algorithm, can then be calculated by summing these probabilities over all pairs, giving^[36] $\sum _{i=2}^{n}\sum _{k=0}^{i-2}{\frac {2}{k+2}}=\sum _{i=1}^{n-1}2H_{i}=O(n\log n).$ The divergence of the harmonic series corresponds in this application to the fact that, in the comparison model of sorting used for quicksort, it is not possible to sort in linear time.^[37]

Deret yang berkaitan

Deret harmonik selang-seling

Empat jumlah parsial pertama dari deret harmonik selang-seling (ruas garis berwarna hitam) memperlihatkan bahwa jumlahnya konvergen menuju ke logaritma alami dari 2 (garis berwarna merah).

Deret $\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots$ dikenal sebagai deret harmonik selang-seling. Deret ini konvergen bersyarat melalui uji deret selang-seling alternating series test, namun tidak konvergen mutlak. Jumlah deretnya sama dengan logaritma alami dari 2.^[38]

Hanya menggunakan pecahan satuan ganjil dengan tanda yang bergantian menghasilkan deret yang berkaitan, yaitu rumus Leibniz untuk $π$ ^[39] $\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots ={\frac {\pi }{4}}.$

Fungsi zeta Riemann

Fungsi zeta Riemann merupakan fungsi yang didefinisikan sebagai deret konvergen untuk bilangan real $x>1$ $\zeta (x)=\sum _{n=1}^{\infty }{\frac {1}{n^{x}}}={\frac {1}{1^{x}}}+{\frac {1}{2^{x}}}+{\frac {1}{3^{x}}}+\cdots ,$ Bila $x=1$ , fungsi zeta Riemann akan berupa deret harmonik. Fungsi ini dapat diperluas melalui kontinuasi analitik dan fungsi holomorfik pada semua bilangan kompleks terkecuali $x=1$ , dimana fungsi yang diperluas ini mempunyai kutub sederhana. Beberapa nilai penting dalam fungsi zeta Riemann, seperti ${\textstyle \zeta (2)={\frac {\pi ^{2}}{6}}}$ yang merupakan nilai untuk solusi masalah Basel, konstanta Apéry $\zeta (3)$ yang merupakan bilangan irasional dibuktikan oleh Roger Apéry, dan "garis kritis" bilangan kompleks dengan bagian real ${\textstyle {\frac {1}{2}}}$ berupa dugaan hipotesis Riemann. Nilai tersebut merupakan satu-satunya nilai selain bilangan negatif agar fungsi tersebut dapat sama dengan nol.^[40]

Deret harmonik acak

Deret harmonik acak ditulis sebagai $\sum _{n=1}^{\infty }{\frac {s_{n}}{n}},$ dimana $s_{n}$ adalah variabel acak terdistribusi independen dan identik yang mengambil dua nilai, yaitu $+1$ dan $-1$ , dengan probabilitas sama dengan ${\textstyle {\frac {1}{2}}}$ . Deret ini konvergen dengan probabilitas 1, yang dapat diperlihatkan dengan menggunakan teorema tiga deret Kolmogerov atau sangat berkaitan dengan pertidaksamaan maksimal Kolmogorov. Jumlah deret ini merupakan variabel acak dengan fungsi kerapatan probabilitas mendekati ke ${\tfrac {1}{4}}$ untuk nilai antara $-1$ dan $1$ , dan deret ini menurun mendekati ke nol untuk nilai yang lebih besar dari $3$ atau lebih kecil dari $-3$ . Jika jumlah deret yang berada di antara nilai $-2$ dan $2$ , maka kerapatan probabilitasnya adalah ${\tfrac {1}{8}}-\varepsilon$ , untuk $\varepsilon$ bilangan taknol, tetapi nilainya sangat kecil, yaitu $\varepsilon <10^{-42}$ .^[41]^[42]

Deret Kempner

Deret Kempner dapat diperlihatkan konvergen ke nilai 22,92067661926415034816...., bila semua suku-suku dengan digit 9 pada penyebut yang muncul dimana-dimana dihilangkan.^[43] Bahkan ketika semua suku memuat suatu untai digit khusus (dalam suatu basis) dihilangkan, deret tersebut konvergen.^[44]

References

^ ^a ^b Rice, Adrian (2011). "The harmonic series: A primer". Dalam Jardine, Dick; Shell-Gellasch, Amy. Mathematical Time Capsules: Historical Modules for the Mathematics Classroom. MAA Notes. 77. Washington, DC: Mathematical Association of America. hlm. 269–276. ISBN 978-0-88385-984-1.
^ ^a ^b ^c ^d Kullman, David E. (May 2001). "What's harmonic about the harmonic series?". The College Mathematics Journal. 32 (3): 201–203. doi:10.2307/2687471. JSTOR 2687471.
^ Hersey, George L. (2001). Architecture and Geometry in the Age of the Baroque. University of Chicago Press. hlm. 11–12, 37–51. ISBN 978-0-226-32783-9.
^ Oresme, Nicole (c. 1360). Quaestiones super Geometriam Euclidis [Questions concerning Euclid's Geometry] (dalam bahasa Latin).
^ Stillwell, John (2010). Mathematics and its History. Undergraduate Texts in Mathematics (edisi ke-3rd). New York: Springer. hlm. 182. doi:10.1007/978-1-4419-6053-5. ISBN 978-1-4419-6052-8. MR 2667826.
^ Derbyshire, John (2003). Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. Washington, DC: Joseph Henry Press. hlm. 10. ISBN 0-309-08549-7. MR 1968857.
^ Mengoli, Pietro (1650). "Praefatio [Preface]". Novae quadraturae arithmeticae, seu De additione fractionum [New arithmetic quadrature (i.e., integration), or On the addition of fractions] (dalam bahasa Latin). Bologna: Giacomo Monti. Mengoli's proof is by contradiction: Let $S$ denote the sum of the series. Group the terms of the series in triplets: $S=1+({\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}})+({\tfrac {1}{5}}+{\tfrac {1}{6}}+{\tfrac {1}{7}})+\cdots$ . Since for $x>1$ , ${\tfrac {1}{x-1}}+{\tfrac {1}{x}}+{\tfrac {1}{x+1}}>{\tfrac {3}{x}}$ , then $S>1+{\tfrac {3}{3}}+{\tfrac {3}{6}}+{\tfrac {3}{9}}+\cdots =1+1+{\tfrac {1}{2}}+{\tfrac {1}{3}}+\cdots =1+S$ , which is impossible for any finite $S$ . Therefore, the series diverges.
^ Bernoulli, Jacob (1689). Propositiones arithmeticae de seriebus infinitis earumque summa finita [Arithmetical propositions about infinite series and their finite sums]. Basel: J. Conrad.
^ Bernoulli, Jacob (1713). Ars conjectandi, opus posthumum. Accedit Tractatus de seriebus infinitis [Theory of inference, posthumous work. With the Treatise on infinite series…]. Basel: Thurneysen. hlm. 250–251.
From p. 250, prop. 16:
"XVI. Summa serei infinita harmonicè progressionalium, ${\tfrac {1}{1}}+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}+{\tfrac {1}{5}}$ &c. est infinita. Id primus deprehendit Frater:…"

[16. The sum of an infinite series of harmonic progression, ${\tfrac {1}{1}}+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}+{\tfrac {1}{5}}+\cdots$ , is infinite. My brother first discovered this…]
^ ^a ^b Dunham, William (January 1987). "The Bernoullis and the harmonic series". The College Mathematics Journal. 18 (1): 18–23. doi:10.1080/07468342.1987.11973001. JSTOR 2686312.
^ Bernoulli, Johann (1742). "Corollary III of De seriebus varia". Opera Omnia. Lausanne & Basel: Marc-Michel Bousquet & Co. vol. 4, p. 8. Johann Bernoulli's proof is also by contradiction. It uses a telescopic sum to represent each term ${\tfrac {1}{n}}$ as ${\frac {1}{n}}={\Big (}{\frac {1}{n}}-{\frac {1}{n+1}}{\Big )}+{\Big (}{\frac {1}{n+1}}-{\frac {1}{n+2}}{\Big )}+{\Big (}{\frac {1}{n+2}}-{\frac {1}{n+3}}{\Big )}\cdots$ $={\frac {1}{n(n+1)}}+{\frac {1}{(n+1)(n+2)}}+{\frac {1}{(n+2)(n+3)}}\cdots$ Changing the order of summation in the corresponding double series gives, in modern notation $S=\sum _{n=1}^{\infty }{\frac {1}{n}}=\sum _{n=1}^{\infty }\sum _{k=n}^{\infty }{\frac {1}{k(k+1)}}=\sum _{k=1}^{\infty }\sum _{n=1}^{k}{\frac {1}{k(k+1)}}$ $=\sum _{k=1}^{\infty }{\frac {k}{k(k+1)}}=\sum _{k=1}^{\infty }{\frac {1}{k+1}}=S-1$ .
^ ^a ^b Knuth, Donald E. (1968). "1.2.7 Harmonic numbers". The Art of Computer Programming, Volume I: Fundamental Algorithms (edisi ke-1st). Addison-Wesley. hlm. 73–78. Knuth writes, of the partial sums of the harmonic series "This sum does not occur very frequently in classical mathematics, and there is no standard notation for it; but in the analysis of algorithms it pops up nearly every time we turn around, and we will consistently use the symbol $H_{n}$ ... The letter $H$ stands for "harmonic", and we call $H_{n}$ a "harmonic number" because [the infinite series] is customarily called the harmonic series."
^ ^a ^b ^c ^d Kifowit, Steven J.; Stamps, Terra A. (Spring 2006). "The harmonic series diverges again and again" (PDF). AMATYC Review. American Mathematical Association of Two-Year Colleges. 27 (2): 31–43. See also unpublished addendum, "More proofs of divergence of the harmonic series" by Kifowit.
^ Roy, Ranjan (December 2007). "Review of A Radical Approach to Real Analysis by David M. Bressoud". SIAM Review. 49 (4): 717–719. JSTOR 20454048. One might point out that Cauchy's condensation test is merely the extension of Oresme's argument for the divergence of the harmonic series
^ ^a ^b Bressoud, David M. (2007). A Radical Approach to Real Analysis. Classroom Resource Materials Series (edisi ke-2nd). Washington, DC: Mathematical Association of America. hlm. 137–138. ISBN 978-0-88385-747-2. MR 2284828.
^ Boas, R. P. Jr.; Wrench, J. W. Jr. (1971). "Partial sums of the harmonic series". The American Mathematical Monthly. 78: 864–870. doi:10.1080/00029890.1971.11992881. JSTOR 2316476. MR 0289994.
^ ^a ^b ^c Havil, Julian (2003). "Chapter 2: The harmonic series". Gamma: Exploring Euler’s Constant. Princeton University Press. hlm. 21–25. ISBN 978-0-691-14133-6.
^ ^a ^b Osler, Thomas J. (November 2012). "96.53 Partial sums of series that cannot be an integer". The Mathematical Gazette. 96 (537): 515–519. doi:10.1017/S0025557200005167. JSTOR 24496876. See in particular Theorem 1, p. 516.
^ Sanna, Carlo (2016). "On the $p$ -adic valuation of harmonic numbers". Journal of Number Theory. 166: 41–46. doi:10.1016/j.jnt.2016.02.020. MR 3486261.
^ Ross, Bertram (1978). "The psi function". Mathematics Magazine. 51 (3): 176–179. doi:10.1080/0025570X.1978.11976704. JSTOR 2689999. MR 1572267.
^ ^a ^b Hadley, John; Singmaster, David (March 1992). "Problems to sharpen the young: An annotated translation of Propositiones ad acuendos juvenes". The Mathematical Gazette. 76 (475): 102–126. doi:10.2307/3620384. JSTOR 3620384. See problem 52: De homine patrefamilias – A lord of the manor, pp. 124–125.
^ Gale, David (May 1970). "The jeep once more or jeeper by the dozen". The American Mathematical Monthly. 77 (5): 493–501. doi:10.1080/00029890.1970.11992525. JSTOR 2317382.
^ Graham, Ronald; Knuth, Donald E.; Patashnik, Oren (1989). "6.3 Harmonic numbers". Concrete Mathematics (edisi ke-2e). Addison-Wesley. hlm. 272–278. ISBN 978-0-201-55802-9.
^ ^a ^b Sharp, R. T. (1954). "Problem 52: Overhanging dominoes" (PDF). Pi Mu Epsilon Journal. 1 (10): 411–412.
^ Paterson, Mike; Peres, Yuval; Thorup, Mikkel; Winkler, Peter; Zwick, Uri (2009). "Maximum overhang". The American Mathematical Monthly. 116 (9): 763–787. doi:10.4169/000298909X474855. MR 2572086.
^ Euler, Leonhard (1737). "Variae observationes circa series infinitas" [Various observations concerning infinite series]. Commentarii Academiae Scientiarum Petropolitanae (dalam bahasa Latin). 9: 160–188.
^ ^a ^b Rubinstein-Salzedo, Simon (2017). "Could Euler have conjectured the prime number theorem?". Mathematics Magazine. 90 (5): 355–359. doi:10.4169/math.mag.90.5.355. JSTOR 10.4169/math.mag.90.5.355. MR 3738242.
^ Pollack, Paul (2015). "Euler and the partial sums of the prime harmonic series". Elemente der Mathematik. 70 (1): 13–20. doi:10.4171/EM/268. MR 3300350.
^ Tsang, Kai-Man (2010). "Recent progress on the Dirichlet divisor problem and the mean square of the Riemann zeta-function". Science China. 53 (9): 2561–2572. doi:10.1007/s11425-010-4068-6. MR 2718848.
^ Maunsell, F. G. (October 1938). "A problem in cartophily". The Mathematical Gazette. 22 (251): 328–331. doi:10.2307/3607889. JSTOR 3607889.
^ Gerke, Oke (April 2013). "How much is it going to cost me to complete a collection of football trading cards?". Teaching Statistics. 35 (2): 89–93. doi:10.1111/test.12005.
^ Parker, Matt (February 12, 2022). "The coupon collector's problem (with Geoff Marshall)". Stand-up maths. YouTube.
^ Luko, Stephen N. (March 2009). "The "coupon collector's problem" and quality control". Quality Engineering. 21 (2): 168–181. doi:10.1080/08982110802642555.
^ Frieze, Alan; Karoński, Michał (2016). "4.1 Connectivity". Introduction to Random Graphs. Cambridge University Press, Cambridge. hlm. 64–68. doi:10.1017/CBO9781316339831. ISBN 978-1-107-11850-8. MR 3675279.
^ Isaac, Richard (1995). "8.4 The coupon collector's problem solved". The Pleasures of Probability. Undergraduate Texts in Mathematics. New York: Springer-Verlag. hlm. 80–82. doi:10.1007/978-1-4612-0819-8. ISBN 0-387-94415-X. MR 1329545.
^ Templat:Introduction to Algorithms
^ Cormen et al. (2009), Section 8.1, "Lower bounds for sorting", pp. 191–193.
^ Freniche, Francisco J. (2010). "On Riemann's rearrangement theorem for the alternating harmonic series". The American Mathematical Monthly. 117 (5): 442–448. doi:10.4169/000298910X485969. JSTOR 10.4169/000298910x485969. MR 2663251.
^ Soddy, F. (1943). "The three infinite harmonic series and their sums (with topical reference to the Newton and Leibniz series for $\pi$ )". Proceedings of the Royal Society. 182: 113–129. doi:10.1098/rspa.1943.0026. MR 0009207.
^ Bombieri, E. (2010). "The classical theory of zeta and $L$ -functions". Milan Journal of Mathematics. 78 (1): 11–59. doi:10.1007/s00032-010-0121-8. MR 2684771.
^ Schmuland, Byron (May 2003). "Random harmonic series" (PDF). The American Mathematical Monthly. 110 (5): 407–416. doi:10.2307/3647827. JSTOR 3647827.
^ Bettin, Sandro; Molteni, Giuseppe; Sanna, Carlo (2018). "Small values of signed harmonic sums". Comptes Rendus Mathématique. 356 (11-12): 1062–1074. arXiv:1806.05402  . doi:10.1016/j.crma.2018.11.007. hdl:2434/634047. MR 3907571.
^ Baillie, Robert (May 1979). "Sums of reciprocals of integers missing a given digit". The American Mathematical Monthly. 86 (5): 372–374. doi:10.1080/00029890.1979.11994810. JSTOR 2321096.
^ Schmelzer, Thomas; Baillie, Robert (June 2008). "Summing a curious, slowly convergent series". The American Mathematical Monthly. 115 (6): 545–540. JSTOR 27642532.

External links

(Inggris) Weisstein, Eric W. "Harmonic Series". MathWorld.

Templat:Series (mathematics)

[rice-1] Rice, Adrian (2011). "The harmonic series: A primer". Dalam Jardine, Dick; Shell-Gellasch, Amy. Mathematical Time Capsules: Historical Modules for the Mathematics Classroom. MAA Notes. 77. Washington, DC: Mathematical Association of America. hlm. 269–276. ISBN 978-0-88385-984-1.

[kullman-2] Kullman, David E. (May 2001). "What's harmonic about the harmonic series?". The College Mathematics Journal. 32 (3): 201–203. doi:10.2307/2687471. JSTOR 2687471.

[hersey-3] Hersey, George L. (2001). Architecture and Geometry in the Age of the Baroque. University of Chicago Press. hlm. 11–12, 37–51. ISBN 978-0-226-32783-9.

[oresme-4] Oresme, Nicole (c. 1360). Quaestiones super Geometriam Euclidis [Questions concerning Euclid's Geometry] (dalam bahasa Latin).

[stillwell-5] Stillwell, John (2010). Mathematics and its History. Undergraduate Texts in Mathematics (edisi ke-3rd). New York: Springer. hlm. 182. doi:10.1007/978-1-4419-6053-5. ISBN 978-1-4419-6052-8. MR 2667826.

[derbyshire-6] Derbyshire, John (2003). Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. Washington, DC: Joseph Henry Press. hlm. 10. ISBN 0-309-08549-7. MR 1968857.

[mengoli-7] Mengoli, Pietro (1650). "Praefatio [Preface]". Novae quadraturae arithmeticae, seu De additione fractionum [New arithmetic quadrature (i.e., integration), or On the addition of fractions] (dalam bahasa Latin). Bologna: Giacomo Monti. Mengoli's proof is by contradiction: Let $S$ denote the sum of the series. Group the terms of the series in triplets: $S=1+({\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}})+({\tfrac {1}{5}}+{\tfrac {1}{6}}+{\tfrac {1}{7}})+\cdots$ . Since for $x>1$ , ${\tfrac {1}{x-1}}+{\tfrac {1}{x}}+{\tfrac {1}{x+1}}>{\tfrac {3}{x}}$ , then $S>1+{\tfrac {3}{3}}+{\tfrac {3}{6}}+{\tfrac {3}{9}}+\cdots =1+1+{\tfrac {1}{2}}+{\tfrac {1}{3}}+\cdots =1+S$ , which is impossible for any finite $S$ . Therefore, the series diverges.

[jacob1-8] Bernoulli, Jacob (1689). Propositiones arithmeticae de seriebus infinitis earumque summa finita [Arithmetical propositions about infinite series and their finite sums]. Basel: J. Conrad.

[jacob2-9] Bernoulli, Jacob (1713). Ars conjectandi, opus posthumum. Accedit Tractatus de seriebus infinitis [Theory of inference, posthumous work. With the Treatise on infinite series…]. Basel: Thurneysen. hlm. 250–251.
From p. 250, prop. 16:
"XVI. Summa serei infinita harmonicè progressionalium, ${\tfrac {1}{1}}+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}+{\tfrac {1}{5}}$ &c. est infinita. Id primus deprehendit Frater:…"

[16. The sum of an infinite series of harmonic progression, ${\tfrac {1}{1}}+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}+{\tfrac {1}{5}}+\cdots$ , is infinite. My brother first discovered this…]

[dunham-10] Dunham, William (January 1987). "The Bernoullis and the harmonic series". The College Mathematics Journal. 18 (1): 18–23. doi:10.1080/07468342.1987.11973001. JSTOR 2686312.

[johann-11] Bernoulli, Johann (1742). "Corollary III of De seriebus varia". Opera Omnia. Lausanne & Basel: Marc-Michel Bousquet & Co. vol. 4, p. 8. Johann Bernoulli's proof is also by contradiction. It uses a telescopic sum to represent each term ${\tfrac {1}{n}}$ as ${\frac {1}{n}}={\Big (}{\frac {1}{n}}-{\frac {1}{n+1}}{\Big )}+{\Big (}{\frac {1}{n+1}}-{\frac {1}{n+2}}{\Big )}+{\Big (}{\frac {1}{n+2}}-{\frac {1}{n+3}}{\Big )}\cdots$ $={\frac {1}{n(n+1)}}+{\frac {1}{(n+1)(n+2)}}+{\frac {1}{(n+2)(n+3)}}\cdots$ Changing the order of summation in the corresponding double series gives, in modern notation $S=\sum _{n=1}^{\infty }{\frac {1}{n}}=\sum _{n=1}^{\infty }\sum _{k=n}^{\infty }{\frac {1}{k(k+1)}}=\sum _{k=1}^{\infty }\sum _{n=1}^{k}{\frac {1}{k(k+1)}}$ $=\sum _{k=1}^{\infty }{\frac {k}{k(k+1)}}=\sum _{k=1}^{\infty }{\frac {1}{k+1}}=S-1$ .

[knuth-12] Knuth, Donald E. (1968). "1.2.7 Harmonic numbers". The Art of Computer Programming, Volume I: Fundamental Algorithms (edisi ke-1st). Addison-Wesley. hlm. 73–78. Knuth writes, of the partial sums of the harmonic series "This sum does not occur very frequently in classical mathematics, and there is no standard notation for it; but in the analysis of algorithms it pops up nearly every time we turn around, and we will consistently use the symbol $H_{n}$ ... The letter $H$ stands for "harmonic", and we call $H_{n}$ a "harmonic number" because [the infinite series] is customarily called the harmonic series."

[kifowit-13] Kifowit, Steven J.; Stamps, Terra A. (Spring 2006). "The harmonic series diverges again and again" (PDF). AMATYC Review. American Mathematical Association of Two-Year Colleges. 27 (2): 31–43. See also unpublished addendum, "More proofs of divergence of the harmonic series" by Kifowit.

[roy-14] Roy, Ranjan (December 2007). "Review of A Radical Approach to Real Analysis by David M. Bressoud". SIAM Review. 49 (4): 717–719. JSTOR 20454048. One might point out that Cauchy's condensation test is merely the extension of Oresme's argument for the divergence of the harmonic series

[bressoud-15] Bressoud, David M. (2007). A Radical Approach to Real Analysis. Classroom Resource Materials Series (edisi ke-2nd). Washington, DC: Mathematical Association of America. hlm. 137–138. ISBN 978-0-88385-747-2. MR 2284828.

[boawre-16] Boas, R. P. Jr.; Wrench, J. W. Jr. (1971). "Partial sums of the harmonic series". The American Mathematical Monthly. 78: 864–870. doi:10.1080/00029890.1971.11992881. JSTOR 2316476. MR 0289994.

[havil-17] Havil, Julian (2003). "Chapter 2: The harmonic series". Gamma: Exploring Euler’s Constant. Princeton University Press. hlm. 21–25. ISBN 978-0-691-14133-6.

[osler-18] Osler, Thomas J. (November 2012). "96.53 Partial sums of series that cannot be an integer". The Mathematical Gazette. 96 (537): 515–519. doi:10.1017/S0025557200005167. JSTOR 24496876. See in particular Theorem 1, p. 516.

[sanna-19] Sanna, Carlo (2016). "On the $p$ -adic valuation of harmonic numbers". Journal of Number Theory. 166: 41–46. doi:10.1016/j.jnt.2016.02.020. MR 3486261.

[ross-20] Ross, Bertram (1978). "The psi function". Mathematics Magazine. 51 (3): 176–179. doi:10.1080/0025570X.1978.11976704. JSTOR 2689999. MR 1572267.

[alcuin-21] Hadley, John; Singmaster, David (March 1992). "Problems to sharpen the young: An annotated translation of Propositiones ad acuendos juvenes". The Mathematical Gazette. 76 (475): 102–126. doi:10.2307/3620384. JSTOR 3620384. See problem 52: De homine patrefamilias – A lord of the manor, pp. 124–125.

[gale-22] Gale, David (May 1970). "The jeep once more or jeeper by the dozen". The American Mathematical Monthly. 77 (5): 493–501. doi:10.1080/00029890.1970.11992525. JSTOR 2317382.

[graham-23] Graham, Ronald; Knuth, Donald E.; Patashnik, Oren (1989). "6.3 Harmonic numbers". Concrete Mathematics (edisi ke-2e). Addison-Wesley. hlm. 272–278. ISBN 978-0-201-55802-9.

[sharp-24] Sharp, R. T. (1954). "Problem 52: Overhanging dominoes" (PDF). Pi Mu Epsilon Journal. 1 (10): 411–412.

[overhang-25] Paterson, Mike; Peres, Yuval; Thorup, Mikkel; Winkler, Peter; Zwick, Uri (2009). "Maximum overhang". The American Mathematical Monthly. 116 (9): 763–787. doi:10.4169/000298909X474855. MR 2572086.

[euler-26] Euler, Leonhard (1737). "Variae observationes circa series infinitas" [Various observations concerning infinite series]. Commentarii Academiae Scientiarum Petropolitanae (dalam bahasa Latin). 9: 160–188.

[rubsal-27] Rubinstein-Salzedo, Simon (2017). "Could Euler have conjectured the prime number theorem?". Mathematics Magazine. 90 (5): 355–359. doi:10.4169/math.mag.90.5.355. JSTOR 10.4169/math.mag.90.5.355. MR 3738242.

[pollack-28] Pollack, Paul (2015). "Euler and the partial sums of the prime harmonic series". Elemente der Mathematik. 70 (1): 13–20. doi:10.4171/EM/268. MR 3300350.

[tsang-29] Tsang, Kai-Man (2010). "Recent progress on the Dirichlet divisor problem and the mean square of the Riemann zeta-function". Science China. 53 (9): 2561–2572. doi:10.1007/s11425-010-4068-6. MR 2718848.

[maunsell-30] Maunsell, F. G. (October 1938). "A problem in cartophily". The Mathematical Gazette. 22 (251): 328–331. doi:10.2307/3607889. JSTOR 3607889.

[gerke-31] Gerke, Oke (April 2013). "How much is it going to cost me to complete a collection of football trading cards?". Teaching Statistics. 35 (2): 89–93. doi:10.1111/test.12005.

[parkrun-32] Parker, Matt (February 12, 2022). "The coupon collector's problem (with Geoff Marshall)". Stand-up maths. YouTube.

[luko-33] Luko, Stephen N. (March 2009). "The "coupon collector's problem" and quality control". Quality Engineering. 21 (2): 168–181. doi:10.1080/08982110802642555.

[frikar-34] Frieze, Alan; Karoński, Michał (2016). "4.1 Connectivity". Introduction to Random Graphs. Cambridge University Press, Cambridge. hlm. 64–68. doi:10.1017/CBO9781316339831. ISBN 978-1-107-11850-8. MR 3675279.

[isaac-35] Isaac, Richard (1995). "8.4 The coupon collector's problem solved". The Pleasures of Probability. Undergraduate Texts in Mathematics. New York: Springer-Verlag. hlm. 80–82. doi:10.1007/978-1-4612-0819-8. ISBN 0-387-94415-X. MR 1329545.

[clrs-36] Templat:Introduction to Algorithms

[FOOTNOTECormenLeisersonRivestStein2009Section_8.1,_"Lower_bounds_for_sorting",_pp._191–193-37] Cormen et al. (2009), Section 8.1, "Lower bounds for sorting", pp. 191–193.

[freniche-38] Freniche, Francisco J. (2010). "On Riemann's rearrangement theorem for the alternating harmonic series". The American Mathematical Monthly. 117 (5): 442–448. doi:10.4169/000298910X485969. JSTOR 10.4169/000298910x485969. MR 2663251.

[soddy-39] Soddy, F. (1943). "The three infinite harmonic series and their sums (with topical reference to the Newton and Leibniz series for $\pi$ )". Proceedings of the Royal Society. 182: 113–129. doi:10.1098/rspa.1943.0026. MR 0009207.

[bombieri-40] Bombieri, E. (2010). "The classical theory of zeta and $L$ -functions". Milan Journal of Mathematics. 78 (1): 11–59. doi:10.1007/s00032-010-0121-8. MR 2684771.

[schmuland-41] Schmuland, Byron (May 2003). "Random harmonic series" (PDF). The American Mathematical Monthly. 110 (5): 407–416. doi:10.2307/3647827. JSTOR 3647827.

[bemosa-42] Bettin, Sandro; Molteni, Giuseppe; Sanna, Carlo (2018). "Small values of signed harmonic sums". Comptes Rendus Mathématique. 356 (11-12): 1062–1074. arXiv:1806.05402  . doi:10.1016/j.crma.2018.11.007. hdl:2434/634047. MR 3907571.

[baillie-43] Baillie, Robert (May 1979). "Sums of reciprocals of integers missing a given digit". The American Mathematical Monthly. 86 (5): 372–374. doi:10.1080/00029890.1979.11994810. JSTOR 2321096.

[schmelzer-44] Schmelzer, Thomas; Baillie, Robert (June 2008). "Summing a curious, slowly convergent series". The American Mathematical Monthly. 115 (6): 545–540. JSTOR 27642532.

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$n$	Jumlah parsial deret harmonik, $H n$
$n$	dibentuk sebagai pecahan		desimal	ukuran relatif
1	1		~1	1
2	3	/2	~1,5	1.5
3	11	/6	~1,83333	1.83333
4	25	/12	~2,08333	2.08333
5	137	/60	~2,28333	2.28333
6	49	/20	~2,45	2.45
7	363	/140	~2,59286	2.59286
8	761	/280	~2,71786	2.71786
9	7129	/2520	~2,82897	2.82897
10	7381	/2520	~2,92897	2.92897
11	83.711	/27.720	~3,01988	3.01988
12	86.021	/27.720	~3,10321	3.10321
13	1.145.993	/360.360	~3,18013	3.18013
14	1.171.733	/360.360	~3,25156	3.25156
15	1.195.757	/360.360	~3,31823	3.31823
16	2.436.559	/720.720	~3,38073	3.38073
17	42.142.223	/12.252.240	~3,43955	3.43955
18	14.274.301	/4.084.080	~3,49511	3.49511
19	275.295.799	/77.597.520	~3,54774	3.54774
20	55.835.135	/15.519.504	~3,59774	3.59774
21	18.858.053	/5.173.168	~3,64536	3.64536
22	19.093.197	/5.173.168	~3,69081	3.69081
23	444.316.699	/118.982.864	~3,73429	3.73429
24	1.347.822.955	/356.948.592	~3,77596	3.77596
25	34.052.522.467	/8.923.714.800	~3,81596	3.81596
26	34.395.742.267	/8.923.714.800	~3,85442	3.85442
27	312.536.252.003	/80.313.433.200	~3,89146	3.89146
28	315.404.588.903	/80.313.433.200	~3,92717	3.92717
29	9.227.046.511.387	/2.329.089.562.800	~3,96165	3.96165
30	9.304.682.830.147	/2.329.089.562.800	~3,99499	3.99499