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Tagged with polylogarithm euler-sums
6
questions
3
votes
1
answer
62
views
Euler Sums of Weight 6
For the past couple of days I have been looking at Euler Sums, and I happened upon this particular one:
$$
\sum_{n=1}^{\infty}\left(-1\right)^{n}\,
\frac{H_{n}}{n^{5}}
$$
I think most people realize ...
- 203
5
votes
3
answers
595
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 ...
- 11.8k
6
votes
1
answer
249
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
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1} = \frac{3 \pi}{...
- 11.8k
6
votes
3
answers
691
views
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?
I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
- 25.8k
14
votes
1
answer
466
views
A peculiar Euler sum
I would like a hand in the computation of the following Euler sum
$$ S=\sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2m+1)(2n+1)^2(2m+2n+1)} \tag{1}$$
which arises from the computation of $\int_{0}^{1}\frac{\...
- 355k
3
votes
1
answer
109
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 ...
- 11.5k