A limit

Evaluate the limit:

$$\lim_{ n \rightarrow +\infty} \left[ \frac{\arctan 1}{n+1} + \frac{\arctan 2}{n+2}+\cdots+\frac{\arctan n }{n+n} \right]$$

Solution

Succesively we have that:

\begin{align*}
  \lim_{n\to + \infty} \sum_{k=1}^n \frac{\arctan k}{n+k} &=\lim_{n\to +\infty} \sum_{k=1}^n \frac{\frac{\pi}{2}-\arctan {\frac{1}{k}}}{n+k} \\
&=\frac{\pi}{2}\lim_{n\to+\infty} \sum_{k=1}^n \frac{1}{n+k}-\lim_{n\to+\infty} \sum_{k=1}^n \frac{\arctan {\frac{1}{k}}}{n+k}
\end{align*}

Well the first limit is just a Riemann sum and evaluates to $\frac{\pi \ln 2}{2}$ while for the second we have that:

 $$0<\lim_{n\to+\infty} \sum_{k=1}^n \frac{\arctan {\frac{1}{k}}}{n+k}<\lim_{n\to+ \infty} \sum_{k=1}^n \frac{1}{k(n+k)}$$

However,

\begin{align*}
 \lim_{n\to+\infty} \sum_{k=1}^n \frac{1}{k(n+k)} &=\lim_{n\to +\infty} \sum_{k=1}^n \frac{1}{n}\left(\frac{1}{k}-\frac{1}{n+k} \right) \\
&=\lim_{n\to +\infty} \frac{2\mathcal{H}_n-\mathcal{H}_{2n}}{n}\\
&=\lim_{n\to +\infty} \frac{\mathcal{O}(\ln n)}{n}\\
&=0
\end{align*}

This means that our initial limit is $\frac{\pi \ln 2}{2}$.