Closed form for $\sum\limits_{n=1}^\infty\frac{W(n^2)}{n^2}$

Question

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 $\alpha\overset{!}{=}\sqrt{2\pi}-\frac{1}{2},ドル but I do not think this is an equality (although it is hard to tell since the series converges so slowly). Using Wolfram Alpha to sum to 10000ドル$ terms, the partial sum appears to be 2ドル.0142453>2.0066283=\sqrt{2\pi}-\frac{1}{2},ドル 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?

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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,ドル 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?

• Jack D'Aurizio: 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},ドル even if not by much.
• Sangchul Lee: 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 $$\sum_{n=1}^{\infty}\frac{W(n^2)}{n^2} = \sqrt{2\pi} - \frac{1}{2} + 2 \int_{1/\sqrt{e}}^{\infty} \frac{\operatorname{Im}W(-t^2+i 0^+)}{(e^{2\pi t} - 1) t^2} ,円 dt. $$ Mathematica 11 seems to be agreeing with this claim that both sides give approx. values 2ドル.016001\cdots$.
• Brevan Ellefsen: Running millions of summations through mathematica gives an answer of about 2ドル.016$
• Sangchul Lee: It just seems to me a pure evil. :s
• Sangchul Lee: @JackD'Aurizio, I take my word back. Honestly I have no idea, but now I am more convinced that it is bounded by $\pi$.

Answer 1

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 $$\frac{1}{\gamma }\sqrt{\Gamma \left(\frac{2}{3}\right)}=S_7+3.775\times 10^{-9}$$