I want to show that there exists a constant $\gamma$ such that:
$$- \frac{\Gamma'(z)}{\Gamma(z)} = \gamma + \frac{1}{z} + \sum_{n=1}^\infty \left(\frac{1}{z+n}-\frac{1}{n} \right)$$
where $\Gamma$ is the Gamma function. This constant is known as the Euler (-Mascheroni) constant, and I would like to prove its existence without directly computing it. Is that possible? We know a few basic properties about the Gamma function, for example:
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$
I got a tip: Find a 1-periodic function and use the functional equation of $\Gamma$ and $\sin$ to find a bound for the growth. All the proofs that I saw where very computation heavy, and I got told that we can prove the existence of such a $\gamma$ without much computation. Can someone give me some advice, what I could try here?
There are multiple ways to prove the Euler–Mascheroni constant exists. Here is one of the typical proofs that works directly from its definition.
We begin with by defining $$ \gamma:=\lim_{n\to\infty}\gamma_n=\lim_{n\to\infty}(H_n-\log n), \tag{1} $$ with $H_n=\sum_{k=1}^n\frac{1}{k}$ being the harmonic numbers. It follows from the integral representation of the natural logarithm $$ \begin{align} \gamma_n &=\sum_{k=1}^n\frac{1}{k}-\int_1^n\frac{1}{x}\,\mathrm dx\\ &=\frac{1}{n}+\sum_{k=1}^{n-1}\frac{1}{k}-\int_1^n\frac{1}{x}\,\mathrm dx\\ &=\frac{1}{n}+\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=1}^{n-1}\int_k^{k+1}\frac{1}{x}\,\mathrm dx\\ &=\frac{1}{n}+\sum_{k=1}^{n-1}\left(\frac{1}{k}-\int_k^{k+1}\frac{1}{x}\,\mathrm dx\right). \end{align} $$ Now make the observation that for some constant $c$ $$ \int_k^{k+1}c\,\mathrm dx=cx\big|_{x=k}^{k+1}=c; $$ hence, $$ \gamma_n =\frac{1}{n}+\sum_{k=1}^{n-1}\int_k^{k+1}\left(\frac{1}{k}-\frac{1}{x}\right)\,\mathrm dx. $$ The integrand $\frac{1}{k}-\frac{1}{x}$ is nonnegative for each $k$ showing that $\gamma_n\geq0$ and therefore if $\gamma$ exists, it must be nonnegative. To establish $\gamma$ exists we just need to prove that $\lim_{n,\to\infty}\gamma_n$ converges. We do this by first writing $$ \begin{align} \int_k^{k+1}\left(\frac{1}{k}-\frac{1}{x}\right)\,\mathrm dx &\leq \int_k^{k+1}\left(\frac{1}{k}-\frac{1}{k+1}\right)\,\mathrm dx\\ &=\frac{1}{k}-\frac{1}{k+1}\\ &=\frac{1}{k(k+1)}\\ &\leq\frac{1}{k^2}. \end{align} $$ So $$ 0\leq\gamma\leq\lim_{n\to\infty}\frac{1}{n}+\sum_{k=1}^{n-1}\frac{1}{k^2}=\frac{\pi^2}{6}<\infty, $$ which completes the proof.