How can one evaluate the following sum :
$$ S=\sum_{n=1}^{\infty} \frac{\ln(n+2)}{n^2} \approx 2.33444 $$
The similar Sum can be given by :
$$ \sum_{n=1}^{\infty} \frac{\ln(n)}{n^2} = -\frac{\pi^2}{6}\left(\,-12\ln(A)+\gamma+ \ln(2)+\ln(\pi) \,\right)$$
I have been unsuccessful with using definitions of the Glaisher Constant $(A)$ to evaluate $S$ , or with integration and directly evaluating the sum $S$.
Q = Is there a closed form for the above Sum $S$ ?
Thank you kindly for your help and time.
EDIT
The Glaisher - Kinkelin constant A is given by the following limit
$$ A= \lim_{n \to \infty } \frac {K(n+1)}{n^{n^2/2+n/2+1/{12}}\exp(\frac{-n^2}{4})} $$
$$ K(n) = \prod_{t=1}^{n-1} t^t $$
Furthermore the following product was found as
$$\prod_{n=1}^{\infty} n^{1/{n^2}} = \left(\frac{A^{12}}{2 \pi e^{\gamma}}\right)^{\pi^2/6}$$
https://en.wikipedia.org/wiki/Glaisher–Kinkelin_constant
You may find the limit also as :
closed form of $\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$
Similarity :
$$ S = \ln \left(\prod_{k=1}^{\infty} (k+2)^{1/k^2}\right) = ? $$
This is a long comment.
This is a "Feynman Trick" inspired method that appears to me illegitimate and which I certainly can't defend, but that nevertheless leads to what appears to be an asymptotic expansion for $S$
Starting with $$S(m)=\sum _{n=1}^{\infty } \frac{\log (n+m)}{n^2}$$ and differentiating under the summation sign $$\sum _{n=1}^{\infty } \left( \frac{\partial } {\partial m}\frac{\log (m+n)}{n^2}\right)=\frac{\pi ^2}{6 m}-\frac{H_m}{ m^2}$$
Using the asymptotic expansion for the Harmonic Number, $H_n$ involving the Bernoulli Numbers, $B_{2k}$
$$H_m \approx \ln m+\gamma+\frac{1}{2m}-\sum _{k=1}^{r} \frac{B_{2 k}}{2 k m^{2 k}}$$
and integrating indefinitely with respect to $m$ and then setting $m=2$ only in the resulting function (avoiding setting a lower limit for the integral, say $m=0$ which is undefined for this function), we obtain the asymptotic approximation given in (1)
$$\sum _{n=1}^{\infty } \frac{\log (n+2)}{n^2} \approx \left(\frac{1}{6} \pi ^2 \log (2)+\frac{\log (2)}{2}+\frac{\gamma }{2}+\frac{1}{2}+\frac{1}{4\ 2^2}-\sum _{k=1}^r \frac{2^{-2 k-1} B_{2 k}}{2 k (2 k+1)}\right)\tag{1}$$