Showing the existence of the Euler-Mascheroni constant 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?
 A: 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.
