# Absolute convergence of $\sum_{n=1}^{\infty} \frac{\sin(2n)}{\sqrt{n}}$

I have to discuss the conditional and absolute convergence of the series: $$\sum_{n=1}^{\infty} \frac{\sin(2n)}{\sqrt{n}}$$ I believe such a series is conditionally convergent but not absolutely convergent, but I lack some evidence.

• Not Absolutely convergent – Empty Jul 11 '15 at 16:44

It is not absolutely convergent. It follows from the equidistribution of the sequence $\{e^{in}\}_{n\in\mathbb{N}}$ in $S^1$, or from the Dirichlet box principle: among three consecutive integer numbers, $\left|\sin(2n)\right|\geq\frac{1}{2}$ holds for at least one of them, hence: $$\sum_{n=1}^{3N}\frac{|\sin(2n)|}{\sqrt{n}}\geq \frac{1}{2}\sum_{i=1}^{N}\frac{1}{\sqrt{3i}}\geq \sqrt{\frac{N}{3}}-\frac{1}{2}.$$ On the other hand, the series is conditionally convergent by Dirichlet's test, since $\frac{1}{\sqrt{n}}$ decreases to zero and the partial sums of $\sin(2n)$ are bounded.

• I have trouble proving that the partial sums of $sin(2n)$ are bounded. – Mihai Jul 11 '15 at 17:11
• @Antanana: search it on MSE. It has been asked before. The point is the $\sin(2n)$ is just the imaginary part of $e^{2in}$, so you end with computing the imaginary part of a geometric sum. – Jack D'Aurizio Jul 11 '15 at 17:13
• Good observation on the sum of the Sin(2n)! Hadn't thought of that. – lulu Jul 11 '15 at 17:53

The absolute series diverges. To see this we just need to argue that Sin(2n) isn't near 0 terribly often, which is the same as asking that 2n not be near a multiple of $\pi$ terribly often. An ungraceful way to see that is to note that if, say, |2n - k$\pi$| < .01 then neither 2n - 2 nor 2n + 2 can be within .01 of any multiple of $\pi$ (as multiples of $\pi$ are separated by something greater than 3). Thus deleting all the terms corresponding to |2n - k$\pi$| < .01 means that, at worst, you are only taking out 1 term in 3. And the surviving terms are each ≥ $\frac{sin(.01)}{\sqrt{n}}$. The comparison now goes in the correct direction.

Note: I did warn that this was ungraceful. I expect it diverges conditionally as well, but do not know for certain.

Another way to show that the series does not converge absolutely is to use

$\displaystyle\frac{\lvert \sin 2n\rvert}{\sqrt{n}}\ge\frac{\sin^2 2n}{\sqrt{n}}=\frac{1}{2}\left(\frac{1-\cos 4n}{\sqrt{n}}\right)=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{\cos 4n}{\sqrt{n}}\right)$,

where $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{\cos 4n}{\sqrt{n}}\right)$ diverges

since $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ diverges and $\displaystyle\sum_{n=1}^{\infty}\frac{\cos 4n}{\sqrt{n}}$ converges by Dirichlet's test.

An idea: You can try Dirichlet test. You have that $\frac{1}{\sqrt{n}}$ converges to zero monotonically, and the only trouble is partial sums of the sine. Here the complex exponential and geometrical series may be of help.

• I should have mentioned that I tried to use the Dirichlet test. The bigges problem I had was the sine. I couldn't bring the series of sine to a nicer form. – Mihai Jul 11 '15 at 17:03
• I will try what you have suggested. – Mihai Jul 11 '15 at 17:03