Simple proof Euler–Mascheroni $\gamma$ constant
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.)
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 $\log(2)$, $$ \begin{align} \gamma &=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\ &\le\sum_{n=1}^\infty\frac1{n(2n+1)}\\ &=\sum_{n=1}^\infty2\left(\frac1{2n}-\frac1{2n+1}\right)\\[6pt] &=2(1-\log(2)) \end{align} $$