Express limit of sum in terms of definite integral

Question

Evaluate the limit by expressing it as a definite integral: \begin{equation} \lim_{n \to \infty} \sum_{k=n+1}^{2n-1} \frac{n}{n^2+k^2} \end{equation}

I'm really confused about tackling this. Although I know it is a Riemann sum, I don't really understand this and am hoping to seek a correct approach to this question. I tried using $t=1/n$ and replacing the limit by \begin{equation} \lim_{t \to 0+} \sum_{k=n+1}^{2n-1} \frac{n^2}{n^2+k^2}t \end{equation} but I don't know what to proceed next.

I would appreciate some help/tips/hints on how to solve this. Many thanks!

Answer 1

$\textbf{Hint}$: Rewrite then reindex the sum as follows

$$\sum_{k=n+1}^{2n-1}\frac{n}{n^2+k^2} = -\frac{1}{2n}+\sum_{k=n}^{2n-1}\frac{n}{n^2+k^2}= -\frac{1}{n}+\sum_{k=0}^{n-1}\frac{n}{n^2+(k+n)^2}$$

The first term vanishes in the limit leaving the last term

$$\sum_{k=0}^{n-1}\frac{1}{1+\left(\frac{k}{n}+1\right)^2}\cdot\frac{1}{n}$$

Can you continue from here?

Answer 2

We can rewrite the sum as $$\sum_{k = n+1}^{2n-1}\frac{n}{n^2 + k^2} = \frac{1}{n}\sum_{k= n+1}^{2n-1} \frac{1}{1 + \left(\frac{k}{n}\right)^2} = -\frac{1}{5n} + \frac{1}{n}\sum_{k= n+1}^{2n} \frac{1}{1 + \left(\frac{k}{n}\right)^2}.$$ I think you can take it from here.

Then taking the limit makes the first summand vanish and the second one turns into the integral $$\int_1^2\frac{1}{1 + x^2}dx = \tan^{-1}(2) - \tan^{-1}(1).$$