# 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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### 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: $$\sum\limits_{1\leq n_1<\cdots<n_k} {\frac{(\pm1)^{n_1}\cdots (\pm1)^{n_k}}{{n_1^{{s_1}} \cdots n_k^{{s_k}}}}}$$ ...
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### 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 ...
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### 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 ...
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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 ...
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### 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 ...
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### 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 ...
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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 $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$ ...
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 ...