I need to find a closed form for the following series where $B_{k+1}$ is the (k+1)th Bernoulli number.
$$\sum_{k=1}^\infty \frac{(-1)^k B_{k+1}}{k+1}$$
Can someone help me to find the closed form of this series: $\sum_{k=1}^\infty \frac{(-1)^k B_{k+1}}{k+1}$?
3
$\begingroup$
sequences-and-series
bernoulli-numbers
-
0Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, [this link](http://meta.math.stackexchange.com/a/9960) might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – 2017-01-21
1 Answers
1
Odd Bernoulli numbers are $0$ aside from $B_1$, so we have
$$\sum_{k=1}^\infty (-1)^k \frac{B_{k+1}}{k+1} = -\sum_{k=1}^\infty \frac{B_{2k}}{2k}$$
but $B_{2n} \sim (-1)^{n-1} 4 \sqrt{\pi n} \left( \frac{n}{\pi e} \right)^{2n}$, so your series won't converge.