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