Find $\lim_{n \to \infty} n^2 \sum^n_{k=1} {1 \over {(n^2+k^2)^2}}$ using Riemann sums

Question

Find $$\lim_{n \to \infty} n^2 \sum^n_{k=1} {1 \over {(n^2+k^2)^2}}$$ using Riemann sums.

I got

$$\lim_{n \to \infty} n^2 \sum^n_{k=1} {1 \over {(n^2+k^2)^2}} = \lim_{n \to \infty} \sum^n_{k=1} {1 \over {n^2}} {1 \over {(1+({k \over n})^2)^2}} $$

Now, this is not the classic $\lim_{n \to \infty} \sum^n_{k=1} {1 \over {n}} {1 \over {1+({k \over n})^2}}$ that I can just define $f(x)={1 \over {1+x^2}}$. The summation is to $n$ and not $n^2$... what should I do?

Answer 1

$$\lim_{n \to \infty} n^2 \sum^n_{k=1} {1 \over {(n^2+k^2)^2}} = \lim_{n \to \infty} \frac{1}{n}\sum^n_{k=1} {1 \over {n}} {1 \over {(1+({k \over n})^2)^2}}=\lim_{n \to \infty} \frac{1}{n}\lim_{n \to \infty}\sum^n_{k=1} {1 \over {n}} {1 \over {(1+({k \over n})^2)^2}}=0\times\int_{0}^1\frac{1}{(1+x^2)^2}=0 $$