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Questions tagged [euler-sums]

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-1
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1answer
83 views

Evaluation of Euler-type sum $\displaystyle\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{(2n+1)^{2}}$ [on hold]

How can one evaluate the sum $\displaystyle\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{(2n+1)^{2}}$? Here $H_{n}$ denotes $n$-th harmonic number.
1
vote
1answer
56 views

Evaluation of Euler - type sum $\sum\limits_{n=1}^{\infty}\frac{\left(H_{2n}-\frac{1}{2}H_{n}\right)^{2}}{n^{2}}$ [on hold]

How can one evaluate the sum $\displaystyle\sum_{n=1}^{\infty}\frac{\left(H_{2n}-\frac{1}{2}H_{n}\right)^{2}}{n^{2}}$? Here $H_{n}$ denotes the $n$-th harmonic number.
0
votes
0answers
30 views

Evaluation of the Euler type sum [duplicate]

How does one evaluate the sum $\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{(2n+1)^{2}}$, where $H_n = \sum_{k=1}^{n} \frac 1k$?
2
votes
3answers
119 views

Can $\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}$, with $H_n$ the $n$-th harmonic number, be written in terms of $\zeta$ values?

The Euler sums are given by $$S_{p,q} = \sum_{n = 1}^{\infty} \frac{H_{n}^{(p)}}{n^q},$$ where $$H_{n}^{(p)} = \sum_{j = 1}^{n} \frac{1}{j^p}.$$ According to Wolfram, Eq. (19), the following special ...
3
votes
0answers
55 views

Alternating series combining harmonic number and zeta values

While evaluating the following fractional part integral, I get stuck on an almost euler sum as highlighted in red colour. Could someone evaluate the red series in terms of well-known constants ? $$\...
0
votes
1answer
56 views

Evaluation of a series involving the harmonic number

Let $\text{H}_n$ denote the n-th harmonic number, does the following sequence have a closed-form ? As an approximation I have got $0.0922514...$ $$ \frac{3}{2}+\lim_{n\to\infty} \bigg(\sum_{k=3}^{n}\...
-1
votes
2answers
36 views

How to prove the following equality involving sums [closed]

I want to prove the following : $\lim\limits_{n\to\infty}\frac{1}{n^a}$$\sum\limits_{k=1}^{n}k^{a-1}$=$\frac{1}{a}$ , $\forall a \geq 1 , a \in N$ Now it is obvious that this holds for $a=1,2,3 ...$ ...
3
votes
0answers
61 views

Sum identity over the reciprocal of central binomial coefficients using their partial fraction decompositions.

Prove $$\sum_{n=1}^\infty \frac{1}{n^k\binom{2n}{n}}=\frac12 {_{k+1}F_k}\left(\underbrace{1,\dots,1}_{k+1};\frac32,\underbrace{2,\dots,2}_{k-1};\frac14\right)$$ where ${_m F_n}$ is the generalized ...
2
votes
1answer
49 views

Generating function of forth powers of harmonic numbers.

Let $x\in (-1,1)$ and let $n\ge 1$ be an integer. Now, let us define a following family of harmonic sums: \begin{eqnarray} S^{(n)}(x):= \sum\limits_{m=1}^\infty [H_m]^n \cdot x^m \end{eqnarray} It is ...
4
votes
1answer
64 views

Incorrect computation of Euler Sum using Complex Residues

First of all, I know this is a long and intimidating question, but I'm sure anyone with a little bit of complex analysis experience can answer it, so please don't turn away. In this paper by Flajolet ...
6
votes
1answer
147 views

What is $\sum\limits_{m=1}^\infty \frac{H_m}{m^6} \cdot (\frac{1}{2})^m$?

We consider a following class of Euler sums: \begin{equation} {\bf H}^{(1)}_p(\frac{1}{2}) := \sum\limits_{m=1}^\infty \frac{H_m}{m^p} \cdot \frac{1}{2^m} \end{equation} Now by using the following ...
6
votes
2answers
91 views

On a certain integral that involves a product of powers of logarithms.

This is a follow-up question to the following questions: Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$ Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$...
2
votes
4answers
112 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{...
0
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0answers
68 views

Doubt in Euler's proof Basel problem

In the proof of the Basel problem Euler reaches a step where expansion of $(\sin x)/x$ is $[1-(x^2/\pi^2)][1-(x^2/4\pi^2)][1-(x^2/9\pi^2)]\cdots$ And from that he got to- $1- x^2\{1/\pi^2 + (1/4\pi^2) ...
4
votes
1answer
55 views

Euler squared sum of order 2 : $\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}$

Let $\mathcal{H}_n$ denote the $n$ - th harmonic number. What techniques would one use to prove that $$\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2} = \zeta^2(3) + \...
3
votes
1answer
90 views

Is there a way to find out a closed form of the sum $\sum_{n=1}^\infty \frac{1}{n^6}\sum_{k=1}^n \frac{H_k}{k}$?

Is there a way to find out a closed form of the sum $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^6}\sum_{k=1}^n \dfrac{H_k}{k}$ ? It appeared to me while calculating $\displaystyle\sum_{n=1}^\infty \...
24
votes
4answers
5k views

Calculating alternating Euler sums of odd powers

Definition $$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$ We define $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$ ...
6
votes
0answers
155 views

A cubic nonlinear Euler sum

Any idea how to solve the following Euler sum $$\sum_{n=1}^\infty \left( \frac{H_n}{n+1}\right)^3 = -\frac{33}{16}\zeta(6)+2\zeta(3)^2$$ I think It can be solved it using contour integration but ...