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 $k=1$.

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    $\begingroup$ The right side is $-(2/\sqrt{5})\log((1+\sqrt{5})/2)$, which is $-{\rm Res}_{s=1}\zeta_{\mathbf Q(\sqrt 5)}(s)$. On the left side, $-B_{j+1}/(j+1) = -\zeta(-j)$. Not sure what you can do with that, but suggests you should negate both sides. $\endgroup$ – KConrad Jun 15 '17 at 2:20
  • $\begingroup$ @T.Amdeberhan , are there any typos in this question? Did you check this result numerically? $\endgroup$ – Nemo Jun 15 '17 at 10:06
  • $\begingroup$ @Nemo: Thanks, there was a typo which is corrected now. $\endgroup$ – T. Amdeberhan Jun 15 '17 at 17:13
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    $\begingroup$ Just a curiosity. All Bernoulli polyomials of odd degree have the 0, 1 and 1/2 as "trivial" roots. The only known Bernoulli polyomial of odd degree that has "non-trivial" roots is $B_{11}(x)$, whose "non-trivial" roots are the golden ratio and its conjugate. $\endgroup$ – EFinat-S Jun 16 '17 at 16:57
  • $\begingroup$ Yes, that is rather curious. I'm not sure of its immediate impact here. $\endgroup$ – T. Amdeberhan Jun 16 '17 at 17:07

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}

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    $\begingroup$ What kind of dark magic is this? $\endgroup$ – Vincent Jun 15 '17 at 11:54

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 $$\sum_{k\geq0}\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}=\frac{2\log\phi}{1-2\phi}.$$

  • $\begingroup$ Do you like to check again? $\sum_k(-1)^k\binom{2k}k\frac1{4k+2}$ does not seem to give the desired sum. Am I wrong? $\endgroup$ – Lewi_Sol Jun 15 '17 at 17:29
  • $\begingroup$ Even your sum on the Bernoulli seems wrong, which I checked numerically. $\endgroup$ – Lewi_Sol Jun 15 '17 at 18:05
  • $\begingroup$ Thanks for editing, sorry for the two typos, wrote too fast. $\endgroup$ – Henri Cohen Jun 15 '17 at 20:17
  • $\begingroup$ Can you provide the proofs of the two displays in the "Edit" section? $\endgroup$ – GH from MO Jun 15 '17 at 23:11

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 $\frac{2\log\phi}{1-2\phi}$.


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).

  • $\begingroup$ yes, nice observation $\endgroup$ – Nemo Jun 15 '17 at 10:41

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.


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