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.

Classical Euler Sums are of the form

$$s(p,q)~=~\sum_{n=1}^{\infty}\frac{H_n^{(p)}}{n^q}$$

for integers $p,q$ where $H_n^{(p)}$ is a generalized harmonic number. The value $p+q$ is called the weight of the Euler Sum. Closed-forms are known for $p=1$ and $q\geq2$, for $p=q$ and $p+q\geq4$, for $p+q\geq5$ and $q\geq2$ and for some exceptional cases.

For $p=1$ and $q\geq2$ we have Euler's formula

$$\sum_{n=1}^\infty\frac{H_n}{n^q}~=~\left(1+\frac q2\right)\zeta(q+1)+\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

For $p=q$ and $p+q\geq4$ we have

$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^p}~=~\frac12\left(\zeta(2p)+\zeta^2(p)\right)$$

Inconsistently, generalizations as attaching extra coefficients, e.g. $2^{-n}$, or an alternating sign $(-1)^n$ are also referred to as Euler Sums.

References

http://mathworld.wolfram.com/EulerSum.html

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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. Can someone provide a nice proof that $$A(1,1) =…
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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 formula above , but how ? Do there exist other proofs…
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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 $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$ Note the alternating general…
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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…
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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 $(1)$ as similar (namely taking $x$ instead of…
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Evaluate $ \sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}^{(3)}}{2n+1}$ and $\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{(2n+1)^{3}}$

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…
Cody
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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) \binom{2n}{n}} \qquad \text{and} \qquad 2. \sum_{n…
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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 different proofs for this result that I could…
omegadot
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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 $\sum_{k = 1}^n 1/k$. $$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$ What I have…
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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\ln^42 \end{equation*} using the identity…
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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 ?…
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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 integral representation: \begin{equation} {\bf…
Przemo
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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 ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$ What is a closed form for…
Przemo
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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 I am interested in solutions using real methods.
Zaid Alyafeai
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Quadratic Euler sums $\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \widetilde H(n)^3}{2 n+1}$

I'm computing a class of harmonic sums. Denote $H(n)=\sum_1^n \frac{1}{k}, \widetilde H(n)=\sum_1^n \frac{(-1)^{k-1}}{k}$ Harmonic numbers, then how to prove $\small\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \widetilde H(n)^2 H(n)}{2 n+1}=\frac{\pi…
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