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.

We have $n(\sqrt{n^{2}+1}-n)=\frac{n(\sqrt{n^{2}+1}-n)(\sqrt{n^{2}+1}+n)}{\sqrt{n^{2}+1}+n}=\frac{n}{\sqrt{n^{2}+1}+n}=\frac{1}{1+\sqrt{1+\frac{1}{n^{2}}}}$; which in the limit of $n\rightarrow \infty$, goes to $\frac{1}{2}$.
Similarly,$\sqrt{n^{2}+1}-n=\frac{1}{\sqrt{n^{2}+1}+n}\rightarrow 0$ as $n\rightarrow \infty$.
