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T. Amdeberhan
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Bernoulli sum meets golden number

Let $B_n$ denote the Bernoulli numbers and let $\phi=\frac{1+\sqrt{5}}2$ be the 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 $k=1$.

T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217