Convergence N'th Harmonic number minus the Natural Logarithm of N. I was hoping if someone could show me the proof of exactly why this converges to the Euler–Mascheroni constant.

The sequence,
$$1+\frac{1}{2}+\cdots +\frac{1}{n}- \ln n $$ converges.
Consider the series
$$\sum\limits_{n=1}^{\infty} \frac{1}{n}-\ln\left( 1+\frac{1}{n} \right).$$
We show that this series converges. We use the inequality,
$$\ln (1+x) < x \qquad \text{for} -1<x <\infty.$$
First, all of its terms are positive, since
$$\ln\left( 1+\frac{1}{n} \right) <\frac{1}{n}.$$
We now make a comparison with the telescoping series, $\sum\limits_{n=1}^{\infty} \frac{1}{n(n+1)}$.
We have,
\begin{equation*}
\begin{split}
-\ln\left( 1+\frac{1}{n} \right)&=\ln \left( \frac{n}{n+1} \right)\\
&=\ln \left( 1-\frac{1}{n+1} \right)\\
&<-\frac{1}{n+1}\\
\end{split}
\end{equation*}
So,
$$\frac{1}{n}-\ln\left( 1+\frac{1}{n} \right) < \frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}.$$
Therefore the series $$\sum\limits_{n=1}^{\infty} \frac{1}{n}-\ln\left( 1+\frac{1}{n} \right)$$
converges.
The $n$th partial sum of this series is
$$1+\frac{1}{2}+\cdots +\frac{1}{n}- \ln (n+1) $$
and this differs from our original sequence by a term of $\ln (n+1) -\ln n= \ln\left( 1+\frac{1}{n} \right)$ but this goes to zero as $n$ goes to infinity.
