Limit of Sum $\lim\limits_{n\to+\infty}n\sum\limits_{k=1}^{n}(n^2+k^2)^{-1}$

Question

Problem: $$\lim_{n\to +\infty} n\left(\sum_{k=1}^{n}\frac{1}{n^2+k^2}\right)=$$ Source: AP Calculus BC

This looks like the definition of a definite integral to me, so I compared it to that equation. $$\int_a^b\!f(x)\,\mathrm{d}x=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_i)\Delta x$$ where $\Delta x=\frac{b-a}{n}$.

However, I am now stuck because $\Delta x$ has $n$ in the denominator whereas the problem has $n$ in the numerator.

Is my approach incorrect or did I apply the definition incorrectly? How do I find the function that is being integrated and $\Delta x$?

Thank you in advance!

Answer 1

HINT: $$\frac n{n^2+k^2}=\dfrac1n\cdot\dfrac1{1+\left(\dfrac kn\right)^2}$$

Use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n g\left(\frac rn\right)=\int_0^1g(x)dx,$$

Answer 2

hint: $\dfrac{n}{n^2+k^2} = \dfrac{1}{n}\dfrac{1}{1+\left(\frac{k}{n}\right)^2}$