# How to show the divergence of $\sum\limits_{n=1}^\infty\frac{\sin(\sqrt{n})}{\sqrt{n}}$

The 10 standard tests taught in class are:

1) $n^{th}$ term test for divergence.(Not applicable: $\lim =0$).

2) Geometric Series(Not applicable).

3) Telescoping Series(Not applicable)

4) Integral Test(Not applicable: $f<0$ sometimes)

5) $p$-series(Not applicable)

6) Direct Comparison(maybe)

7) Limit Comparison(Not applicable $a_n<0$ sometimes)

8) Alternating Series Test(Not Alternating)

9) Ratio Test fails

10) Root Test fails

I did find a hint online that states we should show that for $k^2+1\leq n\leq k^2+k$ we have $\sum\limits_{n=k^2+1}^{k^2+k}\frac{\sin(\sqrt{n})}{\sqrt{n}}>\frac{1}{8}$. Is there an easier way and if not how should we go about showing this?

• Dirichlet test? – user195934 Feb 11 '16 at 15:24

Hint: if you can find $k$ so that $\sqrt{k^2+1}$ is really close to $\dfrac{\pi}{2}$, then for $i$ at most $k$, $\sqrt{k^2+i}$ will also be close to $\dfrac{\pi}{2}$ (Because the difference between $\sqrt{k^2+1}$ and $\sqrt{k^2+i}$ is at most $0.5$, since $(\sqrt{k^2+1}+\dfrac{1}{2})^2>k^2+k$.

Now for such $k$, $\sin(\sqrt{k^2+i})$ for $i$ at most $k$ will be bounded below by some constant $C$(they found something that works with 1/8, but you can find one of your own). Then: $$\sum_{n=k^2+1}^{n=k^2+k} \dfrac{\sin\sqrt n}{\sqrt n} > C \sum_{n=k^2+1}^{n=k^2+k} \dfrac{1}{\sqrt n} > C \sum_{n=k^2+1}^{n=k^2+k} \frac{1}{k} = C$$

• Please use MathJax. – Aaron Maroja Feb 11 '16 at 14:05
• Done, is this okay? – user157036 Feb 11 '16 at 14:16
• It looks a lot better. Take a look at this tutorial though, so you may improve. – Aaron Maroja Feb 11 '16 at 14:18
• I follow your argument, but I'm not seeing why it's sufficient. How do we know there aren't enough negative terms in the series to offset the positive part that diverges. – Tim Raczkowski Feb 11 '16 at 15:33
• I've done some numerical testing with this sum and it seems to me that it diverges because the sum oscillates. The suggested solution here indicates that the sum would be infinite. So we have: $S_{10} = 3.50$, $S_{100} = 3.16$, $S_{1000} = -.44$, $S_{10000} = -.21$, $S_{100000} = 2.47$, etc. – JEM Feb 11 '16 at 15:57

The hint you stated can't be true for all $k$ but it gives an idea on how to show the serie is divergent.

First remember that on $[2k\pi+\pi/4;2k\pi+3\pi/4]$ (for $k$ an integer) $\sin(x)\geq \sqrt 2 /2$. Now the condition $\sqrt n \in [2k\pi+\pi/4;2k\pi+3\pi/4]$ is equivalent to $n\in [4k^2\pi^2+k\pi^2+\pi^2/16;4k^2\pi^2+3k\pi^2+9\pi^2/16]$, and this last interval has a length of $2k\pi^2+\pi^2/2$, which is greater than $18k+2$ (using the fact that $\pi\geq3)$. So this interval contains at least $18k$ integers.

Thus we have

$$\sum_{2k\pi+\pi/4\leq \sqrt n\leq2k\pi+3\pi/4}\frac{\sin(\sqrt n)}{\sqrt n}\geq \sum_{2k\pi+\pi/4\leq \sqrt n\leq2k\pi+3\pi/4}\frac{\sqrt2/2}{ {2k\pi+3\pi/4}}\geq 18k\frac{\sqrt2}{2\cdot ( {2k\pi+3\pi/4})}$$

which is greater than some (strictly positive) constant.

So the serie must be divergent.

I dont think there is an (significantly) easier way to prove this result though.