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.
 A: 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 $n\rightarrow \infty$.
A: About estimating $\sqrt {n^2 + 1},$ for $n \geq 1$ we get
$$  \left( n + \frac{1}{2n} - \frac{1}{8 n^3} \right)^2 < n^2 + 1 < \left( n + \frac{1}{2n} \right)^2 $$
which you can confirm by simply multiplying out the squares involved. So,
$$  n + \frac{1}{2n} - \frac{1}{8 n^3} < \sqrt{n^2 + 1} <  n + \frac{1}{2n}  $$
I did this by algebra, but there is no problem taking several terms in the Taylor series for $\sqrt {1+t}$ and then substituting $t = \frac{1}{n^2}$
A: Hint:
Rationalise  multiplying and dividing by the conjugate expression.
A: 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
$$\lim_{n \to \infty} f(n) = \frac{1}{2}.$$
