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Apologies if this has been asked before. I am wondering if the following series has a closed form:

$$\alpha=\sum_{n=1}^\infty\frac{W(n^2)}{n^2}\tag{1}$$

where $W(x)$ is the Lambert W function. I am interested in this series as an extension of the series:

$$\sum_{n=1}^\infty\frac{\ln{n^2}}{n^2}=-2\zeta'(2)=\frac{\pi^2}{3}\ln{\left(\frac{A^{12}}{2\pi e^\gamma}\right)}\tag{2}$$

since $W(x)$ is the product logarithm. Obviously $(1)$ must converge by comparison with $(2)$ since $W(x)e^{W(x)}=x$ so $W(x)e^{W(x)}<\ln(x)e^{\ln{x}}$ so $W(x)<\ln{x}$ for $x\ge1$ by monotonicity of $W(x)$. However, I am not sure what it converges to. A value (obtained by a couple of toy formal methods) is $\alpha\overset{!}{=}\sqrt{2\pi}-\frac{1}{2}$, but I do not think this is an equality (although it is hard to tell since the series converges so slowly). Using Wolfram Alpha to sum to $10000$ terms, the partial sum appears to be $2.0142453>2.0066283=\sqrt{2\pi}-\frac{1}{2}$, but I do not know $\alpha$'s exact numerical value.

Thus my question is: Is there a closed form for $\alpha$? If not, is there an expression for the error in the $\sqrt{2\pi}-\frac{1}{2}$ approximation?

My attempts: (Note: I have only put the following here to show where I got the value of $\sqrt{2\pi}-\frac{1}{2}$ from; I assume that the methods are not properly correct). In the first place, using this toy resummation formula I had derived, the formula $\int_0^\infty\frac{W(x^2)}{x^2}dx=\sqrt{2\pi}$ (derivable from $\int_0^\infty\frac{W(x)}{x\sqrt{x}}dx=\sqrt{8\pi}$) and the Taylor series of $W(x)$ I formally calculated $\alpha=\sqrt{2\pi}-\frac{1}{2}$. However, the given resummation formula often gives the wrong answer.

I also attempted to evaluate the integral using the Abel-Plana formula (although I do not think $\frac{W(z^2)}{z^2}$ satisfies the conditions to apply it). Formally using the fact that $\lim\limits_{s\rightarrow0}\frac{W(s^2)}{s^2}=1$, I got:

$$\sum_{n=1}^\infty\frac{W(n^2)}{n^2}=\int_{0}^\infty\frac{W(x^2)}{x^2}\;dx-\frac{1}{2}+i\int_0^\infty\frac{\frac{W(-t^2)}{-t^2}-\frac{W(-t^2)}{-t^2}}{e^{2\pi t}-1}\;dt=\sqrt{2\pi}-\frac{1}{2}$$

I assume that the occurrences of this incorrect value are to do with my incorrect applications of these rules missing some remainder term, but I do not know a good way of rectifying this. So my question is: does anyone know a way of evaluating $(*)$ exactly?

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Anon
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  • What makes you think this has as closed form?? – Brevan Ellefsen Feb 08 '17 at 03:01
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    This is a really strange question, but a closed form might exist. For instance, $\frac{W(n)}{n^2}$ is the residue at $z=W(n)$ of $\frac{\pi(1+z)\cot(\pi z e^z)}{z e^z}$. However, I do not think the hypothesis of EMC or Abel-Plana are met, and numerically $\sum_{n\geq 1}\frac{W(n^2)}{n^2}$ is a bit larger than $\sqrt{2\pi}-\frac{1}{2}$, even if not by much. – Jack D'Aurizio Feb 08 '17 at 03:04
  • @BrevanEllefsen I don't know that it does, but I'm wondering because it was easy to attack by the methods I tried (since the corresponding infinite integral is easy), but which seemed to give the wrong answer. Because of this I was wondering whether there was a better way to approach it which would give a result. I was also led to speculate by comparison with the series involving the usual logarithm, since I know that $W(x)$ sometimes does lend itself to fairly simple analyses due to its definition. I have been wondering about the value of this series for quite a long time now. – Anon Feb 08 '17 at 03:06
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    The issue is that $W(z^2)/z^2$ has branch cut along the imaginary axis. I guess that resolving this branch-cut issue yields $$\sum_{n=1}^{\infty}\frac{W(n^2)}{n^2} = \sqrt{2\pi} - \frac{1}{2} + 2 \int_{1/\sqrt{e}}^{\infty} \frac{\operatorname{Im}W(-t^2+i 0^+)}{(e^{2\pi t} - 1) t^2} , dt. $$ Mathematica 11 seems to be agreeing with this claim that both sides give approx. values $2.016001\cdots$. – Sangchul Lee Feb 08 '17 at 03:09
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    Running millions of summations through mathematica gives an answer of about $2.016$ – Brevan Ellefsen Feb 08 '17 at 03:09
  • @SangchulLee The integral looks quite evil. – Anon Feb 08 '17 at 03:10
  • @SangchulLee nice observation. Any idea how to integrate such a beast? – Brevan Ellefsen Feb 08 '17 at 03:11
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    It just seems to me a pure evil. :s – Sangchul Lee Feb 08 '17 at 03:12
  • @SangchulLee: is, by chance, $\text{Im}W(-t^2)$ bounded by $\pi$ for $t>\exp(-1/2)$? This would lead to a reasonable approximation of the error term. – Jack D'Aurizio Feb 08 '17 at 03:17
  • @SangchulLee: I have to check. I said that because I thought that the most reasonable way for solving $xe^x=-M$ is to apply Newton's method with starting point $x=\frac{\log M}{\log\log M}+\pi i$. – Jack D'Aurizio Feb 08 '17 at 03:32
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    @JackD'Aurizio, I take my word back. Honestly I have no idea, but now I am more convinced that it is bounded by $\pi$. – Sangchul Lee Feb 08 '17 at 03:34
  • Me too, but a better starting point should be $\log M-\log\log M+\pi i$. – Jack D'Aurizio Feb 08 '17 at 03:38
  • @JackD'Aurizio, It seems to me that $f(z) = ze^z$ maps $$\mathcal{D} = {x+iy \in \Bbb{C} : y \in (-\pi, \pi) \text{ and } x > -y \cot y }$$ injectively onto $\Bbb{C}\setminus(-\infty, -e^{-1}]$. That is, $W$ is a map $\Bbb{C}\setminus(-\infty, -e^{-1}] \to \mathcal{D}$. – Sangchul Lee Feb 08 '17 at 04:51
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    You sure line $(2)$ shouldn't read $$\sum_{n=1}^\infty\frac{\ln(n^2)}{n^2}=-2\zeta'(2)?$$ – Simply Beautiful Art Feb 15 '17 at 14:44
  • @SimplyBeautifulArt That's certainly what I meant, thanks for pointing that out. – Anon Feb 15 '17 at 22:41
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    I don't know if can be useful, but note that from a direct application of the Mellin transform it is possible to prove that $$\sum_{n\geq1}\frac{W(n^{2})}{n^{2}}=\sqrt{2\pi}-\frac{1}{2}+\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(2s+2)(-s)^{-s}\frac{\Gamma(s)}{s}ds$$ where $c<-1.$ – Marco Cantarini Feb 19 '17 at 12:03
  • @MarcoCantarini I'm not sure exactly how you proved it, but it looks interesting. I'm not sure where to go from that though... It seems vaguely similar to the form of Ramanujan's master theorem. – Anon Feb 21 '17 at 00:42
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    @Anon I used a property of the Mellin transform. If you want I can write my calculations. Note that the integral doesn't vanish when $c \rightarrow - \infty$ and for this reason the closed form is not only $\sqrt{2 \pi}-\frac{1}{2}.$ – Marco Cantarini Feb 21 '17 at 08:03
  • @MarcoCantarini Thank you, that's alright. I can't really see a nice way of evaluating that integral anyway, although it is a nice way of writing the error term. – Anon Feb 23 '17 at 03:07
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    Maybe Poisson summation formula works, but also this $W(x)$ series gives you a series of $\sum\limits_{n=1}^\infty \frac{\ln^{m-k}(2\ln(n))}{\ln^{m-1}(n)n^2}$. This $n$ sum can use Stirling numbers to evaluate it. – Тyma Gaidash Jun 13 '23 at 22:16

1 Answers1

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Just a few numbers

Computing $$S_k=\sum_{n=1}^{10^k}\frac{W\left(n^2\right)}{n^2}$$ $$\left( \begin{array}{cc} k & S_k \\ 3 & 2.00276605784206527342788 \\ 4 & 2.01424531490397345666972 \\ 5 & 2.01578185484283675359672 \\ 6 & 2.01597490197032815619980 \\ 7 & 2.01599818252359384501875 \\ \end{array} \right)$$

The last value seems to be quite close to $$\frac{1}{\gamma }\sqrt{\Gamma \left(\frac{2}{3}\right)}=S_7+3.775\times 10^{-9}$$

Claude Leibovici
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