<s>[INST] 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 ?
 [/INST] Note that
$$\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \sum_{k=1}^n \frac{2k}{k^2+n^2+1} - \frac{2}{2+n^2} - \frac{4}{5+n^2}.$$
We can ignore the last two terms since they converge to $0$.
Consider
$$\sum_{k=1}^n \frac{2k}{k^2+n^2+1} = \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/n^2)}. $$
This is almost a Riemann sum for $\int_0^1 2x/(1 +x^2) \, dx$ except for the annoying term $1/n^2$.
However, it is valid to take a limit of a double sequence
$$S_{mn} =\frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/m^2)}, $$
as
$$\lim_{n \to \infty}S_{nn} = \lim_{n \to \infty} \lim_{m \to \infty} S_{mn},$$
since the inner limit on the RHS exhibits uniform convergence.
Thus,
$$\lim_{n \to \infty}\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \lim_{n \to \infty} \lim_{m \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/m^2)} \\ = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2} \\ = \int_0^1 \frac{2x}{1+ x^2} \, dx  \\ = \log 2$$
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