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 $k=1$.
 A: 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
$$\sum_{k\geq0}\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}=\frac{2\log\phi}{1-2\phi}.$$
A: Here is a short proof for the 2nd formula in T. Amdeberhan's edit to H. Cohen' post.
The key point is to observe that
\begin{equation*}
 \frac{1}{\binom{2k}{k}(4k+2)} = \frac12 B(k+1,k+1),
\end{equation*}
where $B(\cdot,\cdot)$ is the beta function. With this observation we obtain
\begin{equation*}
\frac12\sum_{k\ge 0}(-1)^{k-1}B(k+1,k+1) = \frac12\int_0^1 \Bigl(\sum_{k\ge 0}(-1)^{k-1}t^k(1-t)^k\Bigr)dt = \frac12\int_0^1 \frac{-1}{1+t-t^2}dt
\end{equation*}
which is easily verified to equal $\frac{2\log\phi}{1-2\phi}$.
A: There is a trivial misprint in the question: the sum should start at k=0 and
not at k=1. Numerically the result is then perfect. Now for the proof... !!!
I tried to replace the Bernoulli numbers by Bernoulli polynomials $B_n(x)$:
observations:
1) The sum wildly diverges if $x$ does not belong to a very small set of
values (even if $x$ is extremely close to $0$). This should be easy.
2) For $x = 1$, trivially works the result is as for $x=0$ plus $1$ by
trivial property of $B_n(1)$.
3) For $x=1/2$ the series converges also very fast, the result is as for
$x=0$ plus $2/5$.
4) For $x=3/2$ the series converges also result plus $4+2/5$. 
I would venture that the series converges if and only if $x$ is an integer
or a half integer (the values can then be easily proved as soon as the initial
identity is proved).
A: 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.
A: Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation
$$
\sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\cdot\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+(1+t^2)^2}=0.069591059035995961110566767049...
$$
So the alternative form of the question is
$$
\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+(1+t^2)^2}=\frac14+\frac{\ln\phi}{1-2\phi}
$$
$\it{Proof}.$ Since
$$
\frac{1}{t^2+(1+t^2)^2}=\frac{1}{\sqrt{5}}\left(\frac1{t^2+1/\phi^2}-\frac1{t^2+\phi^2}\right),
$$
and according to Binet's second integral representation for the digamma function $\psi$
$$
\psi(\phi)=-\frac1{2\phi}+\ln\phi-2\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+\phi^2},
$$
$$
\psi(1/\phi)=-\frac{\phi}{2}-\ln\phi-2\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+1/\phi^2},
$$
and 
$$
\psi(\phi)-\psi(1/\phi)=\frac1{\phi-1}
$$
one has 
\begin{align}
&\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+(1+t^2)^2}\\
&=\frac1{2\sqrt5}\left(\psi(\phi)-\psi(1/\phi)+\frac1{2\phi}-\frac{\phi}2-2\ln\phi\right)\\
&=\frac14+\frac{\ln\phi}{1-2\phi}.
\end{align}
