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Question:

show that $$I=\sum_{k=1}^{\infty}\dfrac{B_{2k}}{2k(2k-1)}=1-\dfrac{1}{2}\ln{(2\pi)}$$

where $B_{n}$ is Bernoulli number:Bernoulli number

I think we can $$I=\sum_{k=1}^{\infty}\left(\dfrac{1}{2k-1}-\dfrac{1}{2k}\right)B_{2k}$$ then I can't it ,

Thank you

PS: someone ask me,why are you ask some questions: My ansewer is:I'd like share interesting problem.

because I can't comment,hello,Daniel Fischer,why is series is highly divergent?

math110
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    That series is highly divergent. – Daniel Fischer Oct 07 '14 at 14:00
  • Why can't you comment? To answer the question, $\lvert B_{2k}\rvert$ is growing extremely fast. The absolute value of the terms of your series converges to $\infty$. If you intend to use some summation method for divergent series, you should tell which. – Daniel Fischer Oct 07 '14 at 14:23
  • @DanielFischer: Interestingly, the sum of the first few terms are in the right numerical ballpark. That could mean that this should be interpreted in a 'divergent summation' sense, or that the numerator should be something like $(2k)!(2k-1)!$. (The latter option gives results which are within a few percent of $1-\frac{1}{2}\ln(2\pi)$.) So the question is definitely related to a valid one, though it isn't at present. – Semiclassical Oct 07 '14 at 14:33
  • To be more precise, one may explicitly calculate $\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}t^{2k}=\frac{1}{2}\coth\frac{t}{2}=(e-1)^{-1}-2^{-1}\approx 0.08198$ which is a decent approximation of $1-\frac{1}{2}\ln(2\pi)\approx 0.08106$. – Semiclassical Oct 07 '14 at 14:49

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One method is as follows. By using the integral \begin{align} B_{2n} = (-1)^{n-1} \, 4n \, \int_{0}^{\infty} \frac{ x^{2n-1} \, dx}{e^{2\pi x} -1} \end{align} then the series in question becomes \begin{align} S &= \sum_{n=1}^{\infty} \frac{B_{2n}}{(2n)(2n-1)} = 2 \int_{0}^{\infty} \frac{ \tan^{-1}(x) \, dx}{e^{2\pi x} -1} = 2 \left( \frac{1 + \zeta'(0)}{2} \right) = 1 + \zeta'(0) = 1 - \frac{1}{2} \, \ln(2\pi). \end{align}

Leucippus
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