Prove $\sum_{n=1}^{\infty}(\sqrt{n^2+3}-\sqrt{n^2-1})=\infty$ I want to prove $$\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty$$
If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\neq0$, I can prove it.
In fact, however, $$\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=0$$
As a solution to finding a sum of series, I only know one that uses a partial sum or a progression of differences. How can I apply it to this proof?
I would appreciate it if you could tell me the solution.

Also:
\begin{align}
\sqrt{n^2+1} &= n\left(1+\frac{1}{n^2}\right)^{1/2} = n\left(1+\frac{1}{2n^2} + O(n^{-4})\right)
\\ &= n + \frac{1}{2n} + O(n^{-3})
\\
\sqrt{n^2-3} &= n - \frac{3}{2n} + O(n^{-3})\qquad\text{similarly}
\\
\sqrt{n^2+1} - \sqrt{n^2-3} &= \frac{2}{n} + O(n^{-3})
\end{align}
Thus the series diverges by comparison with $\sum \frac{1}{n}$.

Sorry, I did $+1$ and $-3$ instead of $+3$ and $-1$.
