Questions tagged [euler-sums]

For questions about and related to so-called Euler Sums, that are sums having [tag:harmonic-numbers] and negative integer powers of the index as coefficients.

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2
votes
1answer
86 views

Quadratic Euler Sum $\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}\left(\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}\right)^2$

Question: denote $$J_n=\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}$$ quadratic alternating harmonic number, then how can we calculate $$\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}J_n^2\ \ \text{and}\ \ \sum_{...
2
votes
0answers
32 views

Nonlinear alternating Euler sums

Let $H_{n}^{(s)}=\sum_{k=1}^{n}k^{-s}$ and $\overline{H}_{n}^{(s)}=\sum_{k=1}^{n}(-1)^{k+1}k^{-s}$. Flajolet - Salvy Theorem says that nonlinear Euler sum of the form $$ \sum_{n=1}^{\infty}\frac{H_{n}^...
2
votes
0answers
30 views

Quadratic MZVs $\sum\limits_{1\leq n_1<\cdots<n_k} {\frac{(\pm1)^{n_1}\cdots (\pm1)^{n_k}}{{g_k(n_1)^{{s_1}}\cdots g_k(n_k)^{{s_k}}}}}$

$1$. Definition. Alternating Multiple Zeta Values are defined as sums of form: $$\sum\limits_{1\leq n_1<\cdots<n_k} {\frac{(\pm1)^{n_1}\cdots (\pm1)^{n_k}}{{n_1^{{s_1}} \cdots n_k^{{s_k}}}}}$$ ...
1
vote
0answers
35 views

Formula for the general case of $\sum_{n=1}^\infty \frac{H_n}{n^2} = 2\zeta(3)$ [duplicate]

Introduction I'm reading "Advanced Integration Techniques" by Zaid Alyafeai, and he brings up quite a interesting example regarding Euler sums. The definition is different in the book and on ...
1
vote
0answers
38 views

Relations between $\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$ and alternating Euler sums

Denote $$f(s;s_1,s_2,\ldots,s_k)=\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$$ Can $f(\cdots)$ always be represented as $\mathbb Q$-linear combination of ...
5
votes
2answers
212 views

On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$

My question is: Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$. $$S = \sum_{n = 1}^\...
3
votes
2answers
137 views

Evaluating $\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx$

In an answer to a question found here @user97357329 implies the following integral $$\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx,$$ can be found relatively easily. So far what I ...
5
votes
3answers
155 views

Closed form for the skew-harmonic sum $\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}$

In a post found here it is mentioned that a closed form for the so-called younger brother (younger in the sense the power in the denominator is only squared, rather than cubed as in the linked ...
7
votes
1answer
260 views

An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$

The following classical generalization $$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$ where $\eta(a)=...
5
votes
1answer
89 views

On $\int_0^1\frac{\ln(1-e^{\pi i/3}x)}{e^{-\pi i/3}-x}\ln^3xdx$ and its generalization

Motivation Consider $$I_n=\int_0^1\frac{\ln(1-\omega x)}{\bar\omega-x}\ln^nxdx=\int_0^1\frac{\omega\ln(1-\omega x)}{1-\omega x}\ln^nxdx$$ where $\omega=e^{i\pi/3}$. It is known that $I_0=\frac1{18}\pi^...
0
votes
1answer
23 views

How can one compute power sums with negative exponent?

Is there a general method to find the sum of powers with negative exponent? For example: $\sum_{i = 1}^{N} i^a$ with $ a \in \mathbb{Z} - \mathbb{N}$
8
votes
1answer
266 views

A difficult logarithmic integral and its relation to alternating Euler Sums

The following integral was recently brought up in this thread on AoPS. $$\mathfrak I~=~\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{1+x}\mathrm dx\tag1$$ It is reasonable to ask for a closed-form of ...
0
votes
1answer
74 views

How to prove/disprove that $\sum _{k=1}^{\infty } \frac{(-1)^k \left(H_{k-1}+\log (k)+\gamma \right)}{k}= 0$

Playing around with the harmonic number $H(n) = 1+ \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ and its asymptotic approximation $$H(n\to\infty) \simeq \gamma + \log(n) +\frac{1}{2n} -\frac{1}{12 n^...
4
votes
1answer
134 views

On the alternating Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1}$

In trying to evaluate the integral given here, in a rather circuitous way, I stumbled upon the following alternating Euler sum $$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1} = \frac{3 \pi}{...
2
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0answers
65 views

famous Euler sum [duplicate]

Does anyone know how Euler in the 18th century proved that $$ \sum_{n=1}^{\infty} \frac{H_n}{n^2}=2 \zeta(3) $$ with $H_n$ being the $n$'th harmonic number?
2
votes
2answers
151 views

Euler sum with Bernoulli numbers

In many sources, I find such equality: $$\dfrac{1}{n}\sum_{k=1}^n \binom n k B_kB_{n-k}+B_{n-1}=-B_n$$ where $B_1=-\dfrac{1}{2}$ $$$$However, there don't write how to get it. I think that it's ...
5
votes
3answers
429 views

How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?

I have already evaluated this sum: \begin{equation*} \sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
8
votes
1answer
149 views

Two Euler sums each containing the reciprocal of the central binomial coefficient

Is it possible to find closed-form expressions for the following two Euler sums containing the reciprocal of the central binomial coefficient? $$1. \sum_{n = 0}^\infty \frac{(-1)^n H_n}{(2n + 1) \...
5
votes
2answers
300 views

Evaluate $\int_0^1 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx$

I was playing around with trying to prove the following alternating Euler sum: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3).$$ Here $H_n$ is the Harmonic number. At least two ...
1
vote
1answer
64 views

Closed form for $\sum\limits_{n=1}^{\infty}\frac{O_{n}^{(p)}}{(2n-1)^{q}}$, with $O_{n}^{(s)}=\sum\limits_{k=1}^n\frac1{(2k-1)^{s}}$

Consider the sum $$\sum_{n=1}^{\infty}\frac{O_{n}^{(p)}}{(2n-1)^{q}}\text{, with }O_{n}^{(s)}=1+\frac{1}{3^{s}}+\dots+\frac{1}{(2n-1)^{s}}$$ My question is: if there exists some general theorems ...
3
votes
4answers
233 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 ...
6
votes
2answers
109 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
70 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
54 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
97 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 ...
3
votes
1answer
59 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
76 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
206 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
116 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
5answers
155 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
votes
0answers
109 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
66 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
2answers
126 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 \...
25
votes
4answers
8k 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
1answer
247 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 ...
2
votes
1answer
118 views

Two alternating Euler sums

How can we evaluate the following series: $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right){{\left( {\sum\limits_{k = 1}^...
8
votes
1answer
212 views

Evaluate $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}^{(3)}}{2n+1}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{(2n+1)^{3}}$

I ran into an Euler sum that appears to be rather tough. Apparently, it does indeed have a closed form, so I assume it is doable. May be a false assumption, though. :) $\displaystyle \sum_{n=1}^{\...
74
votes
7answers
3k views

Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

I found the following formula $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$ and it is cited that Euler proved the ...
125
votes
13answers
9k views

Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums. ...