TITLE: How can we show that $\sum_{n=1}^{\infty}\left({x\over n}-\ln{n+x\over n}\right)=x\gamma+\ln(x!)?$
QUESTION [8 upvotes]: A well-known Euler's series representing Euler-Mascheroni's constant
$$
\sum_{n = 1}^{\infty}\left[{1 \over n} -
\ln\left(n + 1 \over n\right)\right] = \gamma\label{1}\tag{1}
$$
The generalisation of \eqref{1} is:

$$
\sum_{n = 1}^{\infty}\left[{x \over n} -
\ln\left(n + x\over n\right)\right] = x\gamma +\ln\left(x!\right)
\label{2}\tag{2}
$$

I have no idea how shows it; it was base on a guess and some numerical trial we conjecture \eqref{2}. How can we prove \eqref{2} ?.

REPLY [5 votes]: The identity follows easily from Weierstrass's definition of the Gamma function:
$$\Gamma(x)={e^{-\gamma x}\over x}\prod_{n=1}^\infty\left(1+{x\over n}\right)^{-1}e^{x/n}$$
Since $x!=x\Gamma(x)$, we have
$$\begin{align}
x\gamma+\ln(x!)
&=x\gamma+\ln(x\Gamma(x))\\
&=x\gamma+\ln x+\ln\Gamma(x)\\
&=x\gamma+\ln x-\gamma x-\ln x+\sum_{n=1}^\infty\left[{x\over n}-\ln\left(1+{x\over n} \right)\right]\\
&=\sum_{n=1}^\infty\left[{x\over n}-\ln\left(1+{x\over n} \right)\right]
\end{align}$$