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
4 Answers
$\begingroup$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.)
$\endgroup$ $\begingroup$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 $\log(2)$,$$ \begin{align} \gamma &=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\ &\le\sum_{n=1}^\infty\frac1{n(2n+1)}\\ &=\sum_{n=1}^\infty2\left(\frac1{2n}-\frac1{2n+1}\right)\\[6pt] &=2(1-\log(2)) \end{align} $$
$\endgroup$ 10 $\begingroup$You can also prove it by drawing the curve of 1/x and then you can draw some rectangles (with lines on $x=1,2,...,n$) and shade them (which will be little above the curve) and find the area of the shaded rectangles that will be the sum $1+1/2+/1/3+...+1/n$ and find the area underneath the curve which will be greater than the area of the shaded rectangles. That way you can prove that the sequence is bounded below. Further to find if it is decreasing or increasing, you can find the value of $s_{n+1}-s{n}$ with the help of same method and it will be lower than 0 so you can conclude that it is decreasing. After that you can easily prove that it is converging to Euler's constant which is $1+1/2+1/3+....+1/n-logn$.
(not an exact answer...just to clear some reasoning)
$\endgroup$ 4 $\begingroup$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.
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