TITLE: Bernoulli sum meets golden number
QUESTION [18 upvotes]: 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 $k=1$.

REPLY [4 votes]: I meant to give a comment (to Henri Cohen's post), but there seemed to be too many characters. The first equation follows from the "reciprocity formula" $$(-1)^{m+1}\sum_{j=0}^k{k\choose j}\frac{B_{m+1+j}}{m+1+j}+(-1)^{k+1}
\sum_{j=0}^m{m\choose j}\frac{B_{k+1+j}}{k+1+j}=\frac{k!m!}{(k+m+1)!}$$quoted in Wikipedia just above here by setting $m=k$. 
The references are 
[$ $1]  M.B. Gelfand, A note on a certain relation among Bernoulli numbers (Russian), Bashkir. Gos. Univ., Uchen. Zap. Ser. Mat. 31 (1968) 215-216.
[2] Takashi Agoh and Karl Dilcher, Reciprocity Relations for Bernoulli Numbers, American Mathematical Monthly, Vol. 115, No.3, (2008), p.237-244.
[3] L. Saalschütz, Verkürtzte Recursionsformeln für die Bernoullischen Zahlen, Zeit. für Math. und Phys. 37 (1892) 374-378.