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

All we use here is the power series expansion for $\log(1+z)$ with $|z|<1$:
$$
\log(1+z)=z-{z^2\over2}+{z^3\over3}-{z^4\over4}+\cdots\tag1
$$
It suffices to note that
$$
\log N=\log{N+1\over1}+\mathcal O\left(\frac1N\right)=\sum_{n\le N}\log{n+1\over n}+\mathcal O\left(\frac1N\right)
$$
Thus, we have
$$
\sum_{n\le N}\frac1n-\log N=\sum_{n\le N}\left\{\frac1n-\log\left(1+\frac1n\right)\right\}+\mathcal O\left(\frac1N\right)\tag2
$$
As $n\to+\infty$, we know from (1) that
$$
\log\left(1+\frac1n\right)=\frac1n+\mathcal O\left(1\over n^2\right)
$$
This is sufficient for us to show that the partial sum on the right hand side of (2) will converge when $N\to+\infty$. Thus, we know the left hand side of (2) converges too.
