Let $B\_n$ denote the [Bernoulli numbers](http://mathworld.wolfram.com/BernoulliNumber.html) and let $\phi=\frac{1+\sqrt{5}}2$ be the [golden ratio](https://en.wikipedia.org/wiki/Golden_ratio).


I encountered the following infinite sum and would like to ask:



> 
> **Question.** Is this true? If so, any proof?
>  $$\sum\_{\pmb{k=0}}^{\infty}\sum\_{j=k}^{2k}\binom{k}{j-k}\frac{B\_{j+1}}{j+1}
> =\frac{2\,\log\phi}{1-2\phi}.$$
> 
> 
> 


*Caveat.* Do not try reversing summations, it diverges! 


**Update.** Thanks to Henri Cohen for observing the typo, the sum has been edited to start at $k=0$. Readers are advised that Nemo's answer is given when the sum begins with $k=1$.



Bernoulli sum meets golden number
OK, in fact this is easy: simply first prove that
$\sum\_{j=k}^{2k}\binom{k}{j-k}\frac{B\_{j+1}}{j+1}=(-1)^k\binom{2k}{k}\frac{1}{4k+2}$, the rest is immediate.


**Edit.** Henri, I'm editing. If you don't agree, please delete it.
We need to prove that
$$\sum\_{j=k}^{2k}\binom{k}{j-k}\frac{B\_{j+1}}{j+1}=\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}.$$
We also need to prove that
$$\sum\_{k\geq0}\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}=\frac{2\log\phi}{1-2\phi}.$$


