How can we show that $\sum_{n=1}^{\infty}\left({x\over n}-\ln{n+x\over n}\right)=x\gamma+\ln(x!)?$

Question

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} ?.

Answer 1

We use the relationship between the Harmonic numbers and the digamma function:

$$\psi(x+1)+\gamma=H_x=\sum_{n=1}^\infty\frac1n-\frac1{n+x}$$

Take the integral of both sides from $0$ to $x$ and you will find that

$$\ln(x!)+\gamma x=\ln(\Gamma(x+1))+\gamma x=\sum_{n=1}^\infty\frac xn-\ln(n+x)+\ln(n)$$

$$\ln(x!)+\gamma x=\sum_{n=1}^\infty\frac xn-\ln\left(\frac{n+x}n\right)$$

Answer 2

You may also use the well-known identity $$\sum_{n\geq2}\frac{\left(-1\right)^{n}x^{n}}{n}\zeta\left(n\right)=\gamma x+\log\left(\Gamma\left(x+1\right)\right),\,-1<x\leq1$$ to find that $$\gamma x+\log\left(\Gamma\left(x+1\right)\right)=\sum_{n\geq2}\frac{\left(-1\right)^{n}x^{n}}{n}\sum_{m\geq1}\frac{1}{m^{n}}=\sum_{m\geq1}\sum_{n\geq2}\frac{\left(-1\right)^{n}}{n}\frac{x^{n}}{m^{n}}$$ $$=\color{red}{\sum_{m\geq1}\left(-\log\left(1+\frac{x}{m}\right)+\frac{x}{m}\right)}.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}\bracks{{x \over n} - \ln\pars{n + x\over n}} = x\gamma +\ln\pars{x!}:\ {\large ?}}$.

\begin{eqnarray} \sum_{n = 1}^{N}\left[{x \over n} - \ln\left(n + x \over n\right)\right] & = & x\left[\sum_{n = 1}^{N}{1 \over n} - \ln\left(N\right)\right] + \left[x\ln\left(N\right) + \ln\left(\prod_{n = 1}^{N}{n \over n + x}\right)\right] \end{eqnarray}

---

The product $\ds{\prod}$ in the above expression becomes: \begin{align} \prod_{n = 1}^{N}{n \over n + x} & = {N! \over \pars{1 + x}^{\overline{N}}} = {N! \over \Gamma\pars{1 + x + N}/\Gamma\pars{1 + x}} = x!\,{N! \over \pars{N + x}!} \\[5mm] &\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, x!\,{\root{2\pi}N^{N + 1/2}\expo{-N} \over \root{2\pi}\pars{N + x}^{N + x + 1/2}\expo{-\pars{N + x}}} = x!\,{N^{N + 1/2}\expo{x} \over N^{N + x + 1/2}\,\pars{1 + x/N}^{N + x + 1/2}} \\[5mm] &\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, x!\,N^{-x}\,{\expo{x} \over \pars{1 + x/N}^{N}} \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,x!\,N^{-x} \end{align}

---

Finally, \begin{align} \sum_{n = 1}^{N}\bracks{{x \over n} - \ln\pars{n + x \over n}} & \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, x\bracks{\sum_{n = 1}^{N}{1 \over n} - \ln\pars{N}} + \bracks{x\ln\pars{N} + \ln\pars{x!\,N^{-x}}} \\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\, \bbx{x\gamma + \ln\pars{x!}} \end{align}

Answer 4

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

$$\begin{align} x\gamma+\ln(x!) &=x\gamma+\ln(x\Gamma(x))\\ &=x\gamma+\ln x+\ln\Gamma(x)\\ &=x\gamma+\ln x-\gamma x-\ln x+\sum_{n=1}^\infty\left[{x\over n}-\ln\left(1+{x\over n} \right)\right]\\ &=\sum_{n=1}^\infty\left[{x\over n}-\ln\left(1+{x\over n} \right)\right] \end{align}$$