0
$\begingroup$

I have to study the values of $p\in\mathbb{R}$ such that $$\sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^p$$ is a convergent series. So far I have found that the main term is equivalent to $n^{-p/2}$, that way $p$ should be $>2$ in order to ensure convergence. Still I cannot use the limit comparison test to prove it, as it does not lead to the constant after dividing.

$\endgroup$
1
  • 1
    $\begingroup$ $$\left(\sqrt{n+1}-\sqrt n\right)^p=\frac1{\left(\sqrt{n+1}+\sqrt n\right)^p}\sim\frac1{2^p}\frac1{n^{p/2}}$$ $\endgroup$ – Did Apr 10 '18 at 13:41
1
$\begingroup$

Convergence for $p>2$ is trivial since $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}$ is bounded between $\frac{1}{2\sqrt{n+1}}$ and $\frac{1}{2\sqrt{n}}$ for any $n\geq 1$, so the $p$-test together with squeezing completely solve the problem. The following identities (courtesy of Ramanujan) might be more challenging to prove: $$ \sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^3 = \frac{3}{2\pi}\sum_{n\geq 1}\frac{1}{n\sqrt{n}}\tag{p=3} $$ $$ \sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^5 = \frac{15}{2\pi^2}\sum_{n\geq 1}\frac{1}{n^2\sqrt{n}}.\tag{p=5} $$ You may have a look at this related question. $$ \sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^4 = \frac{4\pi}{3}\int_{0}^{1}\frac{x^{3/2}(1-x)^{3/2}}{\sin^2(\pi x)}\,dx\tag{p=4} $$ is pretty nice, too.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.