<s>[INST] Convergence of $n \sqrt{n^2+1}-n^2$ I have been asked to study the convergence of the the serie $a_n = n\sqrt{n^2+1}-n^2$. I have been told that the answer is $\frac{1}{2}$ but I am not being able to demonstrate it. I have tried to integrate it and i get that $\int^\infty_1 n\sqrt{n^2+1}-n^2 = \frac{1}{3}[(n^2+1)^{3/2}-n^3]|^\infty_1$. I don't know if any of this is correct or how to follow from here.
I have also read that $\sqrt{n^2+1} - n$ converges to 0 but I don't know how to apply that conclusion here.
 [/INST] Consider using
$$\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{8} + \mathcal{O}(x^{3})$$
for which
\begin{align}
f(n) &= n \, \sqrt{n^{2} + 1} - n^{2} = n^{2} \, \left(\sqrt{1 + \frac{1}{n^{2}}} - 1 \right) \\
&= n^{2} \, \left(\frac{1}{2 \, n^{2}} - \frac{1}{8 \, n^{4}} + \mathcal{O}\left(\frac{1}{n^{6}}\right) \right) \\
&= \frac{1}{2} - \frac{1}{8 \, n^{2}} + \mathcal{O}\left(\frac{1}{n^{4}}\right).
\end{align}
Considering that $n \to \infty$ the limit becomes
$$\lim_{n \to \infty} f(n) = \frac{1}{2}.$$
</s>