I tried to find the answer for the question: Numerical evaluation of $\sum_{N=1}^\infty\left(\frac{1}{\Gamma(N+1)^2}\right)^{\frac{1}{N}}$. I think my result is $4$ times than the expected value. Is this accidental and my solution is not correct? Does my solution have theoretical or some inattention fault?
The solution is:
$S=\sum\limits_{n=1}^\infty\left(\frac{1}{\Gamma^2(n+1)}\right)^{{1}/{n}}=\sum\limits_{n=1}^\infty\left(\frac{1}{n!^2)}\right)^{{1}/{n}}=\sum\limits_{n=1}^\infty \prod\limits_{k=1}^n k^{-2/n}$
$S=\sum\limits_{n=1}^\infty e^{\frac{2}{n}\sum\limits_{k=1}^n \ln(\frac{1}{k})} $
From the exponent we get:
${\frac{2}{n}\sum\limits_{k=1}^n \ln(\frac{1}{k})}={\frac{2}{n}\sum\limits_{k=1}^n \ln(\frac{n}{k} \frac{1}{n})}=\frac{2}{n}\sum\limits_{k=1}^n \ln\frac{n}{k}+\frac{2}{n}\sum\limits_{k=1}^n \ln \frac{1}{n}$
The first term of the exponent is Riemann sum so we get:
$2\int\limits _0^1\ln\frac{1}{x} dx= 2$
Put back into the sum:
$S=\sum\limits_{n=1}^\infty e^{2-\frac{2}{n} n \ln{n}}=\sum\limits_{n=1}^\infty \frac {e^{2}}{n^2}=\zeta(2)e^2 $
At the suggestion of saulspatz I write my comment as an answer.
The "first term of the exponent" is a Riemann sum but leads only to the integral for $n\to\infty$, means: you cannot use it here.