How find this sum $I\sum_{k=1}^{\infty}\frac{B_{2k}}{2k(2k-1)}=1-\frac{1}{2}\ln{(2\pi)}$ 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?

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}
