All Questions
2
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
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