How to prove series convergence: $\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$ I have this series:
$$\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$$
I know that it's convergent (from WolframAlpha) but I need to prove it is convergent. How can I do it?

You have that $$\begin{align*} \dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}&=\frac1n+\frac{(\sqrt{1+n^2}-\sqrt{2+n^2})\cdot(\sqrt{1+n^2}+\sqrt{2+n^2})}{\sqrt{1+n^2}+\sqrt{2+n^2}}=\\\\&=\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\\& \le \frac{1}{n} \end{align*}$$ Then the convergence is established by squaring both sides and using the comparison test.

For the above implication to be correct we also need to show that the middle part is positive. Indeed $$\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\ < \frac{1}{\sqrt{n^2}+\sqrt{n^2}}=\frac{1}{2n}$$ or equivalently $$-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\ >-\frac{1}{2n}$$ which implies that $$\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}>\frac1n-\frac{1}{2n}=\frac1{2n}>0$$ 
