Show that $\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$ Show that 
$$\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$$
How many ways are there to prove it ?
Is there a standard way ?
I was thinking about making it a Riemann sum.
Or telescoping.
What is the easiest way ?
What is the shortest way ?

Observe
\begin{align}
\sum^n_{k=3}\frac{2k}{n^2+k^2+1} = \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+(k/n)^2}
\end{align}
then we have
\begin{align}
\frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k^2n^{-2}}.
\end{align}
Hence it follows
\begin{align}
\int^1_0 \frac{2x}{1+x^2} \ dx=\lim_{n\rightarrow \infty}\frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k^2n^{-2}} = \int^1_0 \frac{2x}{1+x^2}\ dx
\end{align}
which means
\begin{align}
\lim_{n\rightarrow \infty}\sum^n_{k=3} \frac{2k}{1+k^2+n^2}  = \int^1_0 \frac{2x}{1+x^2}\ dx = \log 2.
\end{align}
