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For $a>0$, define $x_n=\dfrac{\sin(\sqrt{n})}{n^a}$.

Prove that $\sum x_n$ converges if and only if $a>\dfrac1 2$.

I proved the convergence when $a>\dfrac1 2$, but I can't derive the other direction. Could someone give me some help? Thanks a lot.

  • 0
    Consider $y_n=\int_n^{n+1}\frac{\sin(\sqrt t)}{t^\alpha}\,dt$ and try to figure out what can be said about the series $\sum y_n$ and $\sum{(x_n-y_n)}$ – 2017-01-29

2 Answers 2

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A straightforward way to show that this series diverges if $a \leqslant 1/2$ is to show that the Cauchy criterion is not satisfied.

If $\sqrt{n} \in [c_1,c_2] = [\pi/6 + 2k\pi, 5\pi/6 + 2k\pi]$ then $\sin \sqrt{n} \geqslant 1/2$ and, with $a \leqslant 1/2$, we have

$$\begin{align} \left|\sum_{c_1 \leqslant \sqrt{n} \leqslant c_2}\frac{\sin \sqrt{n}}{n^a}\right| &= \left|\sum_{c_1^2 \leqslant n \leqslant c_2^2}\frac{\sin \sqrt{n}}{n^a}\right| \\ &\geqslant \frac{1}{2} \frac{c_2^2 - c_1^2}{c_2^{2a}} \\ &\geqslant \frac{1}{2}\frac{c_2^2 - c_1^2}{c_2} \\ &= \frac{1}{2}\frac{\frac{8 \pi^2k}{3} + \frac{2\pi^2}{3}}{\frac{5\pi}{6 }+ 2 k \pi} \end{align}$$

The expression on the right-hand side converges to $2 \pi/3$ as $k \to \infty$. For any $N \in \mathbb{N}$ there exists sufficiently large $k \in \mathbb{N}$ such that $c_2^2 > c_1^2 > N$ and

$$\left|\sum_{c_1^2 \leqslant n \leqslant c_2^2}\frac{\sin \sqrt{n}}{n^a}\right| > \frac{\pi}{3}.$$

Thus the series fails to satisfy the Cauchy criterion and is divergent if $a \leqslant 1/2.$

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This is not easy. Here's an outline:

  • Prove first that the series diverges when $\alpha=\frac 12$.

To that end, form an asymptotic expansion of $e^{i\sqrt{n+1}}-e^{i\sqrt{n}}$ and prove $$e^{i\sqrt{n+1}}-e^{i\sqrt{n}} = \frac{ie^{i\sqrt n}}{2\sqrt n}-\frac{e^{i\sqrt n}}{8n}+O(\frac{1}{n^{3/2}})$$

With real parts, $$\cos (\sqrt{n+1})-\cos(\sqrt{n})=-\frac{\sin(\sqrt n)}{2\sqrt n} - \frac{\cos(\sqrt n)}{8n}+ O(\frac{1}{n^{3/2}})$$

You should be able to prove that $\sum_n \frac{\cos(\sqrt n)}{n}$ converges and $\sum_n O(\frac{1}{n^{3/2}})$ converges. Therefore, $\displaystyle \sum_n \frac{\sin(\sqrt n)}{\sqrt n}$ converges if and only if $\sum_n\cos [(\sqrt{n+1})-\cos(\sqrt{n})]$ converges.

The last series diverges because $\cos (\sqrt{n})$ is a divergent sequence.

  • For $\alpha < \frac 12$, go for contradiction: suppose $\sum_n \frac{\sin(\sqrt n)}{ n^\alpha}$ converges and use partial summation on the series $\sum_n \frac{\sin(\sqrt n)}{\sqrt n}$.