Find $\lim_{n\rightarrow \infty} n \sum ^{2n}_{k=n}\frac{1}{k^2} $.

Question

I am supposed to do this question by defining a partition and then integrating, but I cannot thing of the way to do this. Please give me some steps and then I can apply this to other problems.

Find $\lim_{n\rightarrow \infty} n \sum ^{2n}_{k=n}\frac{1}{k^2} $.

Answer 1

We can write

$$n\sum_{k=n}^{2n}\frac1{k^2}=n\sum_{k=0}^n\frac1{(n+k)^2}=\frac1n\sum_{k=0}^{n}\frac1{\left(1+\frac kn\right)^2}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{(1+x)^2}$$