# What is the sum of Psi/Digamma-function of consecutive arguments? Is there a closed form?

In a consideration of summation of a series $$s = a_0 + a_1 + a_2 + \cdots \tag 1$$ with $$\lim_{k \to \infty} a_k=0$$ but slowly decreasing, the coefficients $a_k$ are somehow related to $1/k^2$ but with some tricky disturbance; but actually I expect that finally $s=\zeta(2)$.

At the moment I observe empirically (based on the first 64 terms) that the coefficients $a_k$ could be nicely separated into the components $$a_k = b_k + r_k \\ \text{ where } \\ b_k= - {\psi(1+k)-\log(1+k) \over 1+k } \tag 2$$ (with $\psi()$ being the digamma-function) and the residuals $r_k$ decreasing rapidly to zero. So if also the series $$B= \sum_{k=0}^\infty b_k \tag 3$$ is convergent and has a closed form I could check the result $s-\zeta(2)$ by the rapidly converging residual series $R=\sum_{k=0}^\infty r_k$

So my question is:

Q1: does the sum $B$ as defined in (3) and (2) have a closed-form expression and if: which?

[Update]
Using W/A I get the interesting powerseries (around $x$ at $\infty$) for each term as $$\begin{eqnarray} {\psi(x)-\ln(x)\over x} &=& \frac{-1}{2x^2}+\frac{-1}{12x^3}+\frac{1}{120x^5} +\frac{-1}{252x^7}+\frac{1}{240x^9}+\frac{-1}{132x^{11}}+\frac{691}{32760x^{13}} + \cdots \\ &=& {\zeta(0)\over x^2}+{\zeta(-1)\over x^3}+{\zeta(-3)\over x^5}+{\zeta(-5)\over x^7}+ \cdots \end{eqnarray}$$ and if that can be summed over consecutive $x$, we had the expression for $B$ as $$B = - (\zeta(0) \cdot \zeta(2)+\zeta(-1) \cdot \zeta(3)+\zeta(-3) \cdot \zeta(5)+ \cdots ) \underset{\mathfrak N}{=} 0.91624014984...$$ where $\mathfrak N$ means Noerlund-summation and I used 64 terms for that zeta-sum. This gives $11$ consistent digits which agree to @Kirill's computation.
Moreover, if I compute inversely $B = \zeta(2)-R$ I arrive at the same value for $B$ to all digits as given by Kirill (except for the sign - I've corrected a sign-error in the above definition of $b_k$ after Kirill had used the original (erroneous) one)

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It would be helpful if the downvoting could be explained and I could see the problem with that question. –  Gottfried Helms Jun 19 '14 at 14:46
Do you have a reason to expect a closed form? It is probably easier to work with $$B = \int_0^1\left(-\frac{\gamma+\log(1-z)}{1-z}+\frac{\ell(z)}{z}\right)\,dz$$ where $\ell(z) = \partial_\mu|_{\mu=0}\mathrm{Li}_\mu(z)$. –  Kirill Jun 19 '14 at 15:07
@Kirill: you are right, I do not actually have a reason that there is a closed-form, I just hoped for it because the summation by the series itself is very slow. It slowly increases above, for instance after 3600 terms, $B_{3600} = -0.916101277030...$ and by that steady (but decreasing) growth it is possibly related to the first derivative $\zeta'(0) =-0.91893853...$ –  Gottfried Helms Jun 19 '14 at 15:19
It is $-0.916240149844295830534809275625733388801447182393876137844189\ldots$, so it's definitely not $\zeta'(0)$. In fact, I'd say the log term is harder than the digamma term. –  Kirill Jun 19 '14 at 15:22
@Kirill: ahh, good to know. Thanks! –  Gottfried Helms Jun 19 '14 at 15:22