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
Best Answer
One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.
It's decreasing because $u_n-u_{n-1} = \frac1n - \log n + \log(n-1) = \frac1n + \log(1-\frac1n) < 0$ for all $n$. (The inequality is valid because $\log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=\frac1n$ yields $\log(1-\frac1n) \le -\frac1n$.)
It's bounded below because $$ \sum_{j=1}^n \frac1j > \int_1^{n+1} \frac{dt}t = \log (n+1) > \log n, $$ and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $\frac1t$ is decreasing.)