<s>[INST] Is there another proof for Euler–Mascheroni Constant? Problem
Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent.
One Proof
This proof is based on the following inequality

$$\frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}$$

where $n=1,2,\cdots$, which will be used repeatedly.
On one hand, we obtain that $$\ln 2-\ln 1<1,~~\ln 3-\ln 2<\frac{1}{2},~~\ln 4-\ln 3<\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n<\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)<1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)>\frac{1}{n+1}>0.$$ This shows that $x_n$ is bounded below.
On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}>0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists.
Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.
 [/INST] Pictorial proof.

In the picture, take $n=11$.  The red graph is $1/x$, so the area under the red graph from $1$ to $n$ is
$$
\int_1^n\frac{dx}{x} = \ln n.
$$  The area of the white rectangles under the graph is
$$
\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
$$
The difference is shown in green,
$$
\text{area}(\text{green}_n) = \ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)
$$
Take the green triangles, translate them to the left, so they are in the strip between $x=0$ and $x=1$.  They are disjoint.  So their areas add to at most the area of the whole strip $(0,1) \times (0,1)$; that is $1$...
$$
\text{area}(\text{green}_n) < 1
$$
Now as $n$ increases, we add more green regions, so this value increases with $n$, and it bounded above by $1$, so it converges and has limit $\le 1$:
$$
0 < \lim_{n\to\infty}\left[\ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)\right] \le 1
$$
The value you want is obtained by subtracting all of this from $1$:
$$
0 \le \lim_{n\to\infty}\left[  \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)- \ln(n)\right] < 1
$$
If we like, we can think of this as the sum of the gray areas.  
These little gray and green regions are approximately triangles... if they were exactly triangles, then our result would be exactly $1/2$.  As it is, $\gamma$ is approximately $1/2$.
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