2
$\begingroup$

This question is based on an answer and comment to this question:

convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$

Does $\displaystyle \lim_{m \to \infty} \sum_{n=1}^m (-1)^n \left[ \sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2 \right] $ exist?

The answers there show that $\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}} \to 2 $, but are not precise enough to show that the difference is monotonic, so the alternating series theorem can not be applied.

$\endgroup$
1
  • 1
    $\begingroup$ Because $1/\sqrt{x}$ is convex function, $1/\sqrt{k}\le\int_{k-1/2}^{k+1/2}\frac{1}{\sqrt{x}}dx$ and here one can make good upper bound of the partial sum, $2(\sqrt{(n+1)^2-1/2}-\sqrt{n^2-1/2})$. $\endgroup$ – didgogns Apr 8 '17 at 1:18
2
$\begingroup$

We can express $2$ as the telescoping sum $$\sum_{k=n^2}^{(n+1)^2-1} (2\sqrt{k+1}-2\sqrt{k}),$$ which lets us rewrite the sum over $n$ as $$\sum_{n=1}^\infty (-1)^n \sum_{k=n^2}^{(n+1)^2-1} \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$ I claim that this series is actually absolutely convergent, for which we just need to prove the convergence of $$\sum_{k=1}^\infty \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$ Consider the inside term: $$\frac1{\sqrt k} - 2\sqrt{k+1} + 2\sqrt k = \frac{2k+1 - 2\sqrt{k(k+1)}}{\sqrt k} = \frac{(\sqrt{k+1} - \sqrt k)^2}{\sqrt k}.$$ We have $$\left(\sqrt k + \frac1{2\sqrt k}\right)^2 = k + 1 + \frac1{4k} > k+1 \implies \sqrt k + \frac1{2\sqrt k} > \sqrt{k+1},$$ so $\sqrt{k+1} - \sqrt k < \frac1{2\sqrt k}$, and therefore $$\sum_{k=1}^\infty \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right) < \sum_{k=1}^\infty \frac1{4k \sqrt k}$$ which converges by the $p$-series test.

$\endgroup$
1
  • $\begingroup$ Also, $\sqrt{k+1} - \sqrt k =\frac1{\sqrt{k+1}+\sqrt k}< \frac1{2\sqrt k}$ $\endgroup$ – didgogns Apr 8 '17 at 2:45

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.