Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ How would you evaluate the following series?
$$\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $$
Thanks.

We have the following important theorem,

Theorem: Let for the monotonic function $f$ ,$\int_{0}^\infty f(x)dx$ exists and we have $\lim_{x\to\infty}f(x)=0$ and $f(x)>0$ then
  we have 
$$\lim_{h\to0^+}h\sum_{v=0}^\infty f(vh)=\int_{0}^\infty f(x)dx$$

It is enough to take $h^{-1}=t$ and $f(x)=\frac{2}{1+x^2}$, then we get the desired result.

So we showed that
  $$\lim_{t\to\infty}\left(\frac{2}{t}+\frac{2t}{t^2+2^2}+\cdots+\frac{2t}{t^2+n^2}+\cdots\right)=\pi$$

Let me show some additional infinite sum by this theorem

We show 
$$\lim_{t\to
 1^{-}}(1-t)^2\left(\frac{t}{1-t}+\frac{2t^2}{1-t^2}+\frac{3t^2}{1-t^2}+\cdots
 \frac{nt^2}{1-t^2}+\cdots\right)=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

Now in previous theorem if we take $e^{-h}=t$ and $f(x)=\frac{xe^{-x}}{1-e^{-x}}$ then since
$$\int_0^\infty\frac{xe^{-x}}{1-e^{-x}}=\int_0^\infty x(\sum_{n=1}^\infty e^{-nx})dx=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
The previous theorem gives us an important relation of infinite sum about Euler constant

We show
$$\lim_{t\to
 1^{-}}(1-t)\left(\frac{t}{1-t}+\frac{t^2}{1-t^2}+\frac{t^3}{1-t^3}+\cdots \frac{t^n}{1-t^n}+\cdots\right)-\log\frac{1}{1-t}=C=\text{Euler
 constant}$$

By using the previous theorem we just need to take $f(x)=\frac{e^{-x}}{1+e^{-x}}$ and $e^{-h}=t$ and since it is known that $$\int_0^\infty e^{-x}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)dx=C=\text{Euler constant}$$
See this book for more examples of G.polya, problems and theorems in analysis , vol  I  Springer 
