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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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    What makes you think this *has* as closed form?? – 2017-02-08
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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. – 2017-02-08
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    @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. – 2017-02-08
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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 $2.016001\cdots$. – 2017-02-08
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    Running millions of summations through mathematica gives an answer of about $2.016$ – 2017-02-08
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    @SangchulLee The integral looks quite evil. – 2017-02-08
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    @SangchulLee nice observation. Any idea *how* to integrate such a beast? – 2017-02-08
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    It just seems to me a pure evil. :s – 2017-02-08
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    @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. – 2017-02-08
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    @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 $x=\frac{\log M}{\log\log M}+\pi i$. – 2017-02-08
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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 $\pi$. – 2017-02-08
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    Me too, but a better starting point should be $\log M-\log\log M+\pi i$. – 2017-02-08
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    @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 $\Bbb{C}\setminus(-\infty, -e^{-1}] \to \mathcal{D}$. – 2017-02-08
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    You sure line $(2)$ shouldn't read $$\sum_{n=1}^\infty\frac{\ln(n^2)}{n^2}=-2\zeta'(2)?$$ – 2017-02-15
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    @SimplyBeautifulArt That's certainly what I meant, thanks for pointing that out. – 2017-02-15
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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 $c<-1.$ – 2017-02-19
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    @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](https://en.wikipedia.org/wiki/Ramanujan's_master_theorem). – 2017-02-21
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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 $\sqrt{2 \pi}-\frac{1}{2}.$ – 2017-02-21
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    @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. – 2017-02-23

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