# Convergence $\sum \frac{\sin \sqrt{n}}{n^{3/2}}, \quad \sum \frac{\sin n}{\sqrt{n}}, \quad\sum \frac{\sin \sqrt{n}}{n^{3/4}}$

I need to examine the convergence of the following infinite series: $$\sum_{n=1}^\infty \frac{\sin \sqrt{n}}{n^{3/2}}, \,\,\,\, \sum_{n=1}^\infty \frac{\sin n}{\sqrt{n}}, \,\,\,\,\sum_{n=1}^\infty \frac{\sin \sqrt{n}}{n^{3/4}}$$

I was able to show that the first two converge. For (1) I used $|\frac{\sin n}{n^{3/2}}| \leq \frac{1}{n^{3/2}}$ and the comparison with a convergent p-series to prove it converges.

For (2) I used the Dirichlet test knowing that $\sum_{n=1}^m \sin n$ is bounded for all $m$ to show it converges.

With the third series, the comparison $|\frac{\sin \sqrt{n}}{n^{3/4}}| \leq \frac{1}{n^{3/4}}$ does not help and I’m pretty sure $\sum_{n=1}^m \sin \sqrt{n}$ is not bounded.

I am unsure how to make progress with this third series.

• Well, you got the first two down. – Simply Beautiful Art Dec 22 '16 at 19:12
• How do you know the partial sums of $\sin n$ are bounded? – Adam Hughes Dec 22 '16 at 19:12
• @AdamHughes math.stackexchange.com/questions/17966/… – Matthew Towers Dec 13 '19 at 14:04
• @MatthewTowers that was rhetorical for the OP to consider, but thanks nonetheless for your good intent :) – Adam Hughes Dec 13 '19 at 18:04

The third series is convergent. While not monotonic, a type of integral test applies.

We can show in general that a series $\sum_{k=1}^\infty f(k)$ and integral $\int_1^\infty f(x) \, dx$ converge and diverge together if $f'$ is absolutely integrable over $[1,\infty).$

The convergence of the series $$\sum_{k=1}^\infty \frac{\sin \sqrt{k}}{k^{3/4}},$$

follows from the convergence of the improper integral

$$\int_1^\infty \frac{\sin \sqrt{x}}{x^{3/4}} \, dx = \int_{1}^{\infty}\frac{2u\sin u}{u^{3/2}} \, du = 2 \int_{1}^{\infty}\frac{\sin u}{\sqrt{u}} \, du,$$

where the RHS integral converges by the Dirichlet test.

Integrating by parts, we see

$$\int_{k-1}^k (x - \lfloor x\rfloor)f'(x) \, dx= \int_{k-1}^k (x - k +1)f'(x) \, dx \\ = \left.(x - k +1)f(x)\right|_{k-1}^k - \int_{k-1}^k f(x) \, dx \\ = f(k) - \int_{k-1}^k f(x) \, dx.$$

Hence,

$$|C_k| := \left|f(k) - \int_{k-1}^kf(x) \, dx\right| \leqslant \int_{k-1}^k\left| (x - \lfloor x \rfloor )f'(x)\right| \, dx \leqslant \int_{k-1}^k|f'(x)| \, dx$$

and we have absolute convergence

$$\sum_{k=2}^\infty|C_k| \leqslant \int_1^\infty |f'(x)| \, dx < \infty.$$

This implies convergence of

$$\sum_{k=2}^\infty C_k = \sum_{k=2}^\infty f(k) - \int_1^\infty f(x) \, dx,$$

whence the series and integral must converge or diverge together.

In this case with $f(x) = \sin \sqrt{x}/x^{3/4}$ we have

$$\int_1^\infty |f'(x)| \, dx = \int_1^\infty \left|\frac{\cos \sqrt{x}}{2x^{5/4}} - \frac{3 \sin \sqrt{x}}{4 x^{7/4}} \right| \, dx < \infty$$

• Oh nice, I did not know the integral test extends like so. – Simply Beautiful Art Dec 22 '16 at 19:15
• @SimplArt: I'm adding a justification above. – RRL Dec 22 '16 at 19:18
• I saw. :D Nice. – Simply Beautiful Art Dec 22 '16 at 19:18
• @RRL May I ask, why/how your absolute convergence implies the convergence of $\sum_{k=2}^\infty f(k) - \int_1^\infty f(x) \, dx$? I have made a post about it, and I still am working on how it is justified ... – UnknownW Nov 16 '18 at 20:30
• @UnknownW: Sure - I'll get back to you shortly. – RRL Nov 16 '18 at 20:41