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.
 A: $\qquad$ 
The best thing is to just plot the graphic of both functions, and to interpret it. Notice how the first blue area on the left is smaller than $1-\dfrac12$, the second is smaller than $\dfrac12-\dfrac13$, the third is smaller than $\dfrac13-\dfrac14$, etc. At the same time, all these quantities are positive, meaning that the series is strictly increasing. But it is also bounded by $1$, since $\bigg(1-\dfrac12\bigg)+\bigg(\dfrac12-\dfrac13\bigg)+\bigg(\dfrac13-\dfrac14\bigg)+\ldots$ is telescopic in nature. And since it is both bounded and monotonous, it follows that it converges.
A: From the error estimate for the rectangle rule:
$$0 < 1/k - \int_k^{k+1} 1/x dx = \frac{1}{2c_k^2}$$
where $c_k \in (k,k+1)$. So
$$0 < 1/k - \int_k^{k+1} 1/x dx < \frac{1}{2k^2}.$$
Therefore
$$\sum_{k=1}^n 1/k - \ln n = \sum_{k=1}^n 1/k - \int_1^n 1/x dx < \sum_{k=1}^n \frac{1}{2k^2} < \sum_{k=1}^\infty \frac{1}{2k^2} = \frac{\pi^2}{12}.$$
Also, this quantity is increasing, so from the monotone convergence theorem there is a limit. This limit is by definition the Euler-Mascheroni constant, and this argument shows that it is less than $\frac{\pi^2}{12}$. One can repeat the argument, using $c_k < (k+1)$ to get a lower bound of $\sum_{k=2}^\infty \frac{1}{2k^2} = \frac{\pi^2}{12}-\frac{1}{2}.$
A: 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.
