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Calculate that$$I=\sum_{k=2}^{\infty}\frac{\ln k}{(k+2)^{3/2}}$$

In fact I haven't dealt with this kind of problem.And, this series converges so slowly that Wolfram alpha do a wrong answer in low digit.

One possible thing I can do is prove that$$I<2+\sqrt{2}.$$But it seems useless.

Bernard
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Vstal
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  • Interesting, Mathematica N[Sum[Log[k]/(k+2)^(3/2),{k,2,Infinity}],80] gives $3.2269874138382259858799993951213658615526991749281667556412273599844657103288563$, while (PARI/GP) sumpos(k=2,log(k)/(k+2)^(3/2)) gives $3.2269874138382259858799993949523024137887834415745074284112130236326655149552720$. They disagree starting from 27th figure. – g.kov May 05 '21 at 19:11
  • @g.kos They are probably using different numerical techniques. Perhaps one is using series acceleration, whereas the other is using Euler-Maclaurin, etc. – K.defaoite May 05 '21 at 19:46
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    If the denominator were simply $k$ instead of $k+2$ the result could be stated in terms of the derivative of the zeta function. As it stands there is no acceptable closed form. – K.defaoite May 05 '21 at 19:52
  • The phrasing seems odd. Does the problem this came from have more information? – RobertTheTutor May 05 '21 at 21:02
  • @K.defaoite I hardly forget this situation. Maybe it really doesn't have a closed form. – Vstal May 06 '21 at 02:12
  • @g.kov. Could you try to look at my last sentence ? I do not understand why I have problems "close" to $a=2$. Thanks. – Claude Leibovici May 06 '21 at 06:54

1 Answers1

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Too long for a comment.

This problem looks very similar to this question.

Consider the more general case

$$I(a)=\sum_{k=2}^{\infty}\frac{\log( k)}{(k+a)^{3/2}}$$ and use a series expansion around $a=0$ $$\frac{\ln k}{(k+a)^{3/2}}=\sum_{n=0}^\infty \binom{-\frac{3}{2}}{n}\, k^{-n-\frac{3}{2}}\log (k)\,\, a^n$$ $$\sum_{k=2}^\infty \binom{-\frac{3}{2}}{n}\, k^{-n-\frac{3}{2}}\log (k)=\frac{2 \sqrt{\pi }}{\Gamma \left(-n-\frac{1}{2}\right) \Gamma (n+1)} \,\zeta '\left(n+\frac{3}{2}\right)$$ making $a=2$ does not gives the correct answer.

For $\color{red}{a=1}$, this works like a charm with a fast convergence. Computing the partial sums (from $n=0$ to $n=p$), we have numerically $$\left( \begin{array}{cc} p & \sum_{n=0}^p \\ 5 & 3.490688272 \\ 10 & 3.499057047 \\ 15 & 3.498737820 \\ 20 & 3.498749643 \\ 25 & 3.498749222 \\ 30 & 3.498749236 \end{array} \right)$$

Same behaviour for any $a =\frac 3 2$ for a correct result.

Could the problem be related to the fact that $$\lim_{n\to \infty } \, \frac{(2 n+3)\,\, \zeta '\left(n+\frac{5}{2}\right)}{(2n+2) \,\, \zeta '\left(n+\frac{3}{2}\right)}=0.5000002500 \quad > \quad \frac 12$$

Claude Leibovici
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