<s>[INST] Evaluating $\lim\limits_{n\to\infty} \sum_{k=1}^{n} \frac{1}{n^2} \cdot \frac{k^2}{\sqrt{n^2+k^2}}$ How would you evaluate the following series?
$$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \frac{1}{n^2} \cdot \frac{k^2}{\sqrt{n^2+k^2}}$$
I tried
$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \frac{1}{n^2} \cdot \frac{k^2}{\sqrt{n^2+k^2}}=
   \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n} \dfrac{\frac{k^2}{n^2}}{\sqrt{1 + \left(\frac{k}{n}\right)^2}}.$
So I have to calculate: $$\int_0^1 \frac{x^2}{\sqrt{1 + x^2}}\,dx?$$
 [/INST] Yes, your integral is correct. You can substitute $x=\tan \theta$ and get the integral (skipping some steps):
$$\int_0^{\pi/4} \left(\frac{1}{\cos^3\theta}-\frac1{\cos \theta}\right)\,d\theta=\\
\int_0^{\pi/4}\left(\frac{1}{(1-\sin^2 \theta)^2}-\frac{1}{1-\sin^2\theta}\right)\cos\theta\,d\theta$$
Letting $u=\sin\theta$ then:
$$\int_0^{\sqrt 2/2} \left(\frac1{(1-u^2)^2}-\frac1{1-u^2}\right)\,du$$
Then apply the standard technique of partial fractions.
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