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

The 10 standard tests taught in class are:

1) nth term test for divergence.(Not applicable: lim).

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?

Answer

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

Attribution
Source : Link , Question Author : JEM , Answer Author : Landon Carter

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