<s>[INST] 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.
 [/INST] $\sqrt{n^2+3}-\sqrt{n^2-1}=$
$=\dfrac{\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\left(\sqrt{n^2+3}+\sqrt{n^2-1}\right)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=$
$=\dfrac{n^2+3-(n^2-1)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=\dfrac4{\sqrt{n^2+3}+\sqrt{n^2-1}}>$
$>\dfrac4{\sqrt{n^2+3n^2}+\sqrt{n^2}}=\dfrac4{\sqrt{4n^2}+\sqrt{n^2}}=$
$=\dfrac4{3n}>\dfrac1n\;,\quad$ for all $\;n\in\mathbb{N}\;.$
Hence,
$\sqrt{n^2+3}-\sqrt{n^2-1}>\dfrac1n\;,\quad$ for all $\;n\in\mathbb{N}\;.$
Since $\;\displaystyle\sum_{n=1}^\infty\dfrac1n=+\infty\;,\;$ by using the comparison test, we get that $\;\displaystyle\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=+\infty\;.$
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