The following interesting sum seems to approach the Euler-Mascheroni constant $\gamma$.
$$\sum_{n=1}^\infty \frac{\ln n!}{n^3} \overset{?}{=} \gamma$$
I've looked at the different ways to express the Euler-Mascheroni constant and tried to apply those methodes to this sum.
I also tried using the series representation of $\displaystyle \frac{\ln n!}{n^3}$ for $\displaystyle n=0,1,2$
and using the log gamma function
$$\displaystyle\ln \Gamma(n+1)=-\gamma-\gamma n-\ln (n+1)+\sum\limits_{k=1}^\infty\frac{n+1}{k}-\ln\left(1+\frac{n+1}{k}\right)$$
This problem might be quite well out of my mathmatical reach but I still would love to know the answer. It would be awesome if anyone could prove or disprove that the sum equals the Euler-Mascheroni constant and show their method.
Best Answer
It is known that for $n\geq 1$, $$ \log n! > \left( {n + \frac{1}{2}} \right)\log n - n + \log \sqrt {2\pi } + \frac{1}{{12n}} - \frac{1}{{360n^3 }}. $$ Thus \begin{align*} \sum\limits_{n = 1}^\infty {\frac{{\log n!}}{{n^3 }}} & > \sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n^2 }}} + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n^3 }}} - \sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}} \\ &\quad + \log \sqrt {2\pi } \sum\limits_{n = 1}^\infty {\frac{1}{{n^3 }}} + \frac{1}{{12}}\sum\limits_{n = 1}^\infty {\frac{1}{{n^4 }}} - \frac{1}{{360}}\sum\limits_{n = 1}^\infty {\frac{1}{{n^6 }}} \\ & = - \zeta '(2) - \frac{1}{2}\zeta '(3) - \zeta (2) + \zeta (3)\log \sqrt {2\pi } + \frac{1}{{12}}\zeta (4) - \frac{1}{{360}}\zeta (6) \\ & = 0.583661 \ldots>\gamma. \end{align*} For the values of the derivative of $\zeta$ I used the OEIS A073002 and A244115. The other values, except $\zeta(3)$, can be expressed in terms of $\pi$.