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Questions tagged [harmonic-numbers]

For questions regarding harmonic numbers, which are partial sums of the harmonic series. The $N$-th harmonic number is the sum of reciprocals of the first $N$ natural numbers.

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On the sums $\sum_{i=1}^{n} \sum_{j=0}^{m-1} \dfrac{a_j}{mi-j}$ as $n \to \infty$

This was inspired by Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ Let $s(n) =\sum_{i=1}^{n} \sum_{j=0}^{m-1} \dfrac{a_j}{mi-j} $ and $...
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98 views

Closed expressions for harmonic-like multiple sums

Inspired by Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $ here are some related problems. Prove that $$w(2,1) = \sum _{i=1}^{\infty } \sum _{j=...
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Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $

First, the original problem follows, $$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}$$ where $$H_{k}=\sum_{j=1}^k \frac{1}{j}$$ is the $k$-th partial sum of harmonic series. Using the following identity, ...
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31 views

Set Consisting of Harmonics of Integer Powers of 2?

If one defines "harmonics" as natural number fractions of a number, is there a name for the set consisting of all harmonics of all integer powers of two?
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116 views

Another “closed form expression” for a generating function involving harmonic numbers.

This question is closely related to Calculating alternating Euler sums of odd powers. Let $p\ge 1$ and $q\ge 1$ be integers and $1> x\ge 0$ be real. We use the following definition: \begin{...
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Generating function of square harmonic numbers

During my study of generating functions, I was able to calculate the generating function of the sequence of harmonic numbers $H_n$: $$\sum_{n=1}^\infty H_nx^n=\frac{\ln(1-x)}{x-1}$$ However, I also ...
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I can't figure out my mistake in calculating the harmonic series

Let's say $a = H_{\infty}$, so $a = \sum_{k=1}^\infty \frac{1}{k}$. $$a = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{\infty}$$ Now if we take $a - a + 1$ we get $$a - a + 1 = 1 + \frac{1}{2} + ...
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98 views

Harmonic number identity $\sum_{k=1}^{n}H_k=(n+1)(H_{n+1}-1)$

Prove using generating functions that for Harmonic numbers $H_n= \sum_{j=1}^{n}\frac{1}{j}$ $$\sum_{k=1}^{n}H_k=(n+1)(H_{n+1}-1) (*)$$ The generating function for harmonic numbers is $$\sum_{n \in \...
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Series whose $a_n \rightarrow 0$ as $n \rightarrow \infty$ and diverges

I was wondering if there are series whose $a_n \rightarrow 0$ as $n \rightarrow \infty$ and diverges. I know one example, that is $a_n = \frac{1}{n}$. Slight modifications don't count, such as $a_n = \...
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42 views

About the partition of some reciprocals

Does there exists a partition of the reciprocals $\frac{1}{2}, \frac{1}{3},\cdots,\frac{1}{12}$, into two sets $A$ and $B$ such that $\displaystyle\sum_A\frac{1}{n} -\sum_B\frac{1}{n}=1$?
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Show that $\sum\limits_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}=3\zeta(3)$, where, for every positive $n$, $H_n=\sum\limits_{k=1}^n\frac1k$

The problem I was considering about is the evaluation of the following series: \begin{align*} \sum_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)} \end{align*} The attempt I could make was to change $(H_{n})^...
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2answers
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A pair of composed integral transforms from Mellin and Laplace transforms

We can create a forward transform $T$ which is a composition of the inverse Mellin $\mathcal{M}^{-1}$ and Laplace $\mathcal{L}$ transforms $$ T[\phi] = \mathcal{L}\mathcal{M}^{-1}[\phi]=\mathcal{L}\...
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1answer
61 views

Tricky partial rational sum

I'm looking for a simplification of $$ \sum _{k=r+1}^{2 r} \frac{2 k+2 r+1}{2 k^2-k (2 r+1)+2 r (r+1)}\:. $$ Mathematica gives a somewhat tautological result in terms of the digamma function $\psi$: $...
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Finding the general/closed form of $\sum_{k=1}^n k^a$ [duplicate]

I recently noticed the following: $$\sum_{i=1}^ni=\frac{n(n+1)}{2}$$ $$\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6} = \Bigg(\sum_{i=1}^ni\Bigg) \cdot \frac{2n+1}{3}$$ $$\sum\limits_{i = 1}^n i^3 = \...
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1answer
53 views

Let $x_n$ denote the smallest integer such that $\sum\limits_{k=n}^{x_n} \frac 1k >1$, then $x_n\sim en$

For $n\geq 2$, let $x_n$ denote the smallest integer such that $\displaystyle \sum_{k=n}^{x_n} \frac 1k >1$. Prove that $\lim_n \frac{x_n}{n}\sim e$. I've found this problem in an old handout of ...
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What is the minimum number of operations needed to calculate exactly the nth Harmonic number

A teacher once told me the minimum number of operations needed to calculate the exact value of the nth Harmonic number was n operations. Using something similar to Euler's proof of: $\sum_{i=1}^n i = \...
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Evaluating $~\sum\limits_{n>0}H(e^n)-n-\gamma~$ in Closed Form

Inspired by these three questions, I asked myself whether $$\sum_{n>0}\Big[~H(e^n)-n-\gamma~\Big]~=~0.278091975548622251874828828459627630\ldots$$ might also possesses a closed form expression, ...
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0answers
118 views

Infinite sum of harmonic number over general polynomial

As an extension to my question Infinite sum of harmonic number over polynomial of 2nd degreee I found a general formula for a polynomial of arbitrary degree which I had presented as a self answer but ...
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Infinite sum of harmonic number over polynomial of 2nd degreee

Inspired by the recent problem How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$ I asked for a generalization, and I found it with ...
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How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+…+\frac{1}{n}\right)$

How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$ Can anyone help me,I really appreciate it.
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2answers
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Prove that: $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\ge\frac{2}{3}$

I've got three inequalities: $\forall n\in\mathbb N:$ $$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{1}{2}$$ $$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{7}{12}$$ $$\frac{1}{n}...
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1answer
56 views

Comparison of sum and integral over squares of Harmonic numbers

This is an extension of the simpler question [1] This time we compare sum and integral over the squares of the harmonic numbers (see [1] for definitions) The sum is $$f_{s2}(n) = \sum_{k=0}^n H_k^2$...
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60 views

Comparison of sum and integral over Harmonic number

The sum of Harmonic numbers $H_n = \sum_{k=1}^n \frac{1}{n}, H_0 = 0$ defined by $$f_s(n) = \sum_{k=0}^n H_k$$ is given by $$f_s(n) = (n+1) H_n - n$$ Now define the integral $$f_i(n) = \int_0^n ...
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Is the function $\frac{\operatorname{Harmonic}(x)}{\operatorname{Airy}(x)}$ a convex function for real numbers $0<x<1$?

For real numbers $0\leq x\leq 1$, I know how to exlore some aspects of the graph ( I was playing with Wolfram Alpha online calculator about its plot, see next paragraph) of the function $$f(x)=\frac{...
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123 views

Exploring the constant $-\int_0^1\log\left(1-H_x\right)\log\left(1+H_{2x}\right)\,dx$, where $H_y$ are harmonic numbers

This morning I try to create interesting integrals involving harmonic numbers. See this Wikipedia. And look at, also if you need it, the definition of the Harmonic number $H_x$ using the digamma ...
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1answer
113 views

Infinite sum involving complete and incomplete zeta function

While playing around with complete and incomplete zeta functions I found a nice formula which to my knowledge was not discussed in MSE. Here it is $$\sum_{k=2}^\infty (\zeta(k)-1) = 1\tag{1}$$ and ...
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Formulas involving the polynomials $\frac{1}{n!}x^n(a_n-b_nx)^n,$ where the coprime integers $a_n,b_n$ satisfy $H_n=\frac{a_n}{b_n}$ for $n\geq 1$

For integers $n\geq 1$ let $$H_n=1+\frac{1}{2}+\ldots+\frac{1}{n}$$ the $nth$ harmonic number. After I've seen the form of the polynomials used by Niven in [1] I wanted to create a puzzle with a new ...
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Partial sum formula for $\sum_{x=1}^n \frac{1}{bx - x^2}$ at specific limit without using digamma function

Given the following sum: $\sum_{x=1}^n \frac{1}{bx - x^2}$ (b is an integer constant) It appears to me that partial sum cannot be calculated without using Harmonic numbers, or Digamma function. ...
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1answer
382 views

A Summability methods which sum the harmonic series

Studying summability theory I've come across many summation methods however by now I know only two not very interesting method which re-sums the harmonic series $\sum_{n=0}^\infty \frac 1{n+1}$ : the ...
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280 views

How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$?

I came across a question asking the value of the following sum: \begin{align} \left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\ +\left(\frac{1}{2}+\frac{1}{3}+\cdots +...
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Is there a closed form for the alternating series of inverse harmonic numbers?

Let $H_n=\sum _{k=1}^n \frac{1}{k}$ be the n-th harmonic number. Since $H_{k+1}>H_k$ for $k=1, 2, 3, ...$ the sequence $\frac{1}{H_k}$ is monotonic decreasing as $n \to \infty$, and the Leibniz ...
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1answer
153 views

Fibonacci and Harmonic

Here is a cute series that I came across and cannot seem to tackle at all. Question: Calculate $$\mathcal{S} = \sum_{n=1}^{\infty} \frac{F_n \mathcal{H}_{n-1}^{(2)}}{n^2 \binom{2n}{n}}$$ where $F_n$...
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1answer
322 views

Is there a closed form for $\sum _{k=1}^n \frac{1}{k}H_{k-1}^2$?

Introduction This question appeared in the study (Sum of powers of Harmonic Numbers) of the sums of fourth powers of harmonic numbers $H_n = \sum _{k=1}^n \frac{1}{k}$ for $n=1,2,3,...$ and $H_0 = 0$....
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244 views

Existence of a sequence $\{\epsilon_n\}_{n\ge 1}$ such that $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} $ converges

I am trying to find a sequence $\{\varepsilon_n\}_{n\ge 1}$ such that $$\lim_{n\to \infty}\varepsilon_n=1~~~~~~~~~~~~~\text{and}~~~~~~~~~\sum_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} <\infty.$$ ...
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Sum of powers of Harmonic Numbers

This is a natural extension of the question Sum of Squares of Harmonic Numbers. I became interested in this question while studying the problem A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\...
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1answer
63 views

How do you calculate the generic formula for $ \sum\limits_{i=0}^{k} \frac 1 {(n-i)}$

For the given series, how do I calculate the value of $$ \sum_{i=0}^{k} \frac 1 {(n-i)}$$ Thanks.
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2answers
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Why does $\sum_{i=1}^x \frac xi - \lfloor \frac xi \rceil \approx 0.03648x$ instead of $x(1-2\gamma)$

Consider the following serie: $\sum_{i=1}^x \lfloor \frac xi \rfloor = \sum_{i=1}^x \sigma(x)$ Where $\sigma(x)$ is the sum of divisors function. This is easy to understand because if you take for ...
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423 views

Why does $(\sum_{i=1}^x \frac{1}{i}) - \ln(x+\frac{1}{2})$ converge to $\gamma$ a lot faster than $(\sum_{i=1}^x \frac{1}{i}) - \ln(x)$

$(\sum_{i=1}^x \frac{1}{i}) -\ln(x)$ 10: 0,62638316097421 100: 0,58220733165153 1000: 0,57771558156821 $(\sum_{i=1}^x \frac{1}{i}) - \ln(x+\frac{1}{2})$ 10: 0,57759299680478 100: 0,...
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Is this sum $\sum_{k=0}^{n}{n\choose k}\sum_{i=0}^{k}(-1)^i{k\choose i}H_{n+k-i}=H_{2n}$ correct?

Given the double sum, which has a closed form of $H_{2n}$ $$\sum_{k=0}^{n}{n\choose k}\sum_{i=0}^{k}(-1)^i{k\choose i}H_{n+k-i}\tag1$$ Where $H_n$ is the Harmonic number $(1)$ is quite an ...
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1answer
46 views

Prove the equalities

I have to fine the following equality $ \sum^{n-1}_{m=0} \frac{1}{(n-m)p} = neH_n$ where $H_n = 1 + 1/2 + 1/3 + .... 1/n$ My Solution: $ \sum^{n-1}_{m=0} \frac{1}{(n-m)p} = \frac{1}{n} + \frac{1}{...
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0answers
100 views

The Harmonic Logarithm and its relation to the Prime Number Theorem

For the purposes of this question, the Harmonic Number, $H_n$ is defined by the finite series sum $H_n=\sum_{k=1}^n\frac{1}{k}$ (n and k being positive integers) and the Harmonic Logarithm, $hlog(n)$ ...
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4answers
833 views

A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2\right]$

The series of squares of harmonic numbers $$ \sum_{n=1}^\infty H_n^2 \tag1 $$ is divergent since $\displaystyle \lim_{n \to \infty} H_n^2 \ne 0$, actually from the classic result (6.3.18), $$ H_n=\ln ...
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3answers
150 views

Inequality $\sum\limits_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$

I recently came across the following exercise: Prove that $$\sum_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$$ for every natural number $n \ge 1$. I immediately tried by induction, by I did not ...
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2answers
715 views

Proving that $~\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{k^2}~H_k~H_{k-1}=\frac{3}{16}~\zeta(4)$

To show that $$\sum\limits_{k=2}^{\infty} \frac{(-1)^{k}}{k^{2}} \, \left(1+\frac{1}{2}+...+\frac{1}{k}\right) \cdot \left(1+\frac{1}{2}+...+\frac{1}{k-1}\right) = \frac{3}{16}\zeta(4).$$ I came ...
2
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0answers
75 views

Is there a simple description for $\sum_{n=1}^{\infty}\frac{1}{p_n} \to \infty$? [duplicate]

Let $p_n$ the $n^\text{th}$ prime number. I mean $p_1=2,p_2=3,p_3=5,...$ now consider this series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{p_n}=\frac12+\frac13+\frac15+\frac17+... $ I know $\...
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3answers
227 views

The limit of $u_n=\sum_{i=1}^n\frac{1}{i}$.

I had this question : Does the sequence $$\begin{align*} u_n=\sum_{i=1}^n\frac{1}{i} \end{align*}$$ have a real end ? Well my teacher said no but the last question was Is $$\begin{align*} u_n=\...
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1answer
65 views

Prove that $\sum_{n=1}^\infty\frac{\mu(n)}{n}H_n\sum_{k=n+1}^\infty\frac{\mu(k)}{k^2}$ is convergent

For integers $n\geq 1$ let $$H_n=1+\frac{1}{2}+\ldots+\frac{1}{n}$$ the $nth$ harmonic number, and $\mu(n)$ the Möbius function. See, if you need it, this Wikipedia to know the definition of Möbius ...
3
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5answers
467 views

Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$

Can someone compute $$ \int_0^1\frac{\ln^4(1+x)\ln x}x \,dx$$ in closed form? I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the ...
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2answers
232 views

Elementary way to calculate the series $\sum\limits_{n=1}^{\infty}\frac{H_n}{n2^n}$

I want to calculate the series of the Basel problem $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n^2}}$ by applying the Euler series transformation. With some effort I got that $$\displaystyle{\frac{\...
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1answer
81 views

Generating function for cubes of Harmonic numbers I

Let us define a following generating function: \begin{equation} {\bf H}^{(1,1,1)}_n(x) := \sum\limits_{m=1}^\infty [H_m]^3 \cdot \frac{x^m}{m^n} \end{equation} Now, by using results from Generating ...