Simple proof Euler–Mascheroni $\gamma$ constant I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges.
At first I want to know if my answer is OK.
My try:
$\lim\limits_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{k} - \log (n)\right) = \lim\limits_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} [\log(k)-\log(k+1)]\right)$
$ = \lim\limits_{n\to\infty} \left(\frac{1}{n} + \sum\limits_{k=1}^{n-1} \left[\log(\frac{k}{k+1})+\frac{1}{k}\right]\right) = \sum\limits_{k=1}^{\infty} \left[\frac{1}{k}-\log(\frac{k+1}{k})\right]$
Now we prove that the last sum converges by the comparison test:
$\frac{1}{k}-\log(\frac{k+1}{k}) < \frac{1}{k^2} \Leftrightarrow k<k^2\log(\frac{k+1}{k})+1$
which surely holds for $k\geqslant 1$

As $ \sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ converges $ \Rightarrow \sum\limits_{k=1}^{\infty} \left[\frac{1}{k}-\log(\frac{k+1}{k})\right]$ converges and we name this limit $\gamma$
q.e.d

Upper Bound
Note that
$$
\begin{align}
\frac1n-\log\left(\frac{n+1}n\right)
&=\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\\
&\le\int_0^{1/n}t\,\mathrm{d}t\\[3pt]
&=\frac1{2n^2}
\end{align}
$$
Therefore,
$$
\begin{align}
\gamma
&=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\
&\le\sum_{n=1}^\infty\frac1{2n^2}\\
&\le\sum_{n=1}^\infty\frac1{2n^2-\frac12}\\
&=\sum_{n=1}^\infty\frac12\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\[9pt]
&=1
\end{align}
$$

Lower Bound
Note that
$$
\begin{align}
\frac1n-\log\left(\frac{n+1}n\right)
&=\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\\
&\ge\int_0^{1/n}\frac{t}{1+\frac1n}\,\mathrm{d}t\\[3pt]
&=\frac1{2n(n+1)}
\end{align}
$$
Therefore,
$$
\begin{align}
\gamma
&=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\
&\ge\sum_{n=1}^\infty\frac1{2n(n+1)}\\[3pt]
&=\sum_{n=1}^\infty\frac12\left(\frac1n-\frac1{n+1}\right)\\[6pt]
&=\frac12
\end{align}
$$

A Better Upper Bound
Using Jensen's Inequality on the concave $\frac{t}{1+t}$, we get
$$
\begin{align}
\frac1n-\log\left(\frac{n+1}n\right)
&=\frac1n\left(n\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\right)\\
&\le\frac1n\frac{n\int_0^{1/n}t\,\mathrm{d}t}{1+n\int_0^{1/n}t\,\mathrm{d}t}\\
&=\frac1{n(2n+1)}
\end{align}
$$
Therefore, since the sum of the Alternating Harmonic Series is $\log(2)$,
$$
\begin{align}
\gamma
&=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\
&\le\sum_{n=1}^\infty\frac1{n(2n+1)}\\
&=\sum_{n=1}^\infty2\left(\frac1{2n}-\frac1{2n+1}\right)\\[6pt]
&=2(1-\log(2))
\end{align}
$$
