# Is $\sum_{n=1}^{\infty} \frac{(-1)^n}{n \sin^2(n)}$ converging?

Statement: let $a_n$ positive sequence such that $\lim_{n \to \infty} a_n=0$.

Prove or disprove that $\sum_{n=1}^{\infty} (-1)^n\cdot{a_n}$ converges.

It's obvious that the series converges if $a_n$ is monotonic (Leibniz), so I defined a sequence which is not monotonic - $a_n=\frac{1}{n \sin^2(n)}$ which is positive and converges to 0 as stated.

I'm trying to decide whether $\sum_{n=1}^{\infty} \frac{(-1)^n}{n \sin^2(n)}$ converges.

I have no clue how to answer it and would appreciate any help.

Thanks!

• Can you find a lower bound for the terms? – Daniel Fischer Apr 26 '14 at 20:10
• $\sin^2(n)$ itself doesn't converge – user88595 Apr 26 '14 at 20:11
• Huh? Did you write it wrong? Intuitively it should be obvious. $\sin^2(n)$ can be made as small as desired by making $n$ larger (for example, $n=314159\approx \pi 10^5$ will have $\sin(n)$ small). So $\sin(n)^{-1}$ can be made as large as possible, as many times as desired, so its sum must diverge. – user18862 Apr 26 '14 at 20:13
• your sequence $a_n=\frac{n}{n\sin ^2(n)}$ doesn't converge to 0. – Badshah Apr 26 '14 at 20:15
• @NeuroFuzzy: The sine of $314159$ does not have small absolute value. – André Nicolas Apr 26 '14 at 20:16

In order to have $\sum_{n=1}^{\infty} a_n$ converge, we would need to have $\lim_{n \to \infty} a_n = 0$. However, note that $$\left|\frac{1}{\sin^2(n)}\right| \geq \frac{1}{\max_{x \in \mathbb{R}} |\sin^2(x)|}=1$$ So, $\lim_{n \to \infty}a_n \neq 0$. So, the sum cannot converge.

This is for the modified version of question about whether following series converges or not: $$\sum_{n=1}^\infty (-1)^n \frac{1}{n\sin^2 n}\tag{*1}$$ The answer is NO because the terms in the series don't converge to $0$ as $n \to \infty$.

Let $\varepsilon$ be any small positive number. Since $\pi$ is irrational, Hurwitz's theorem tells us there are infinitely many pair of integers $(m,n)$, relative prime to each other, satisfying

$$\left|\pi - \frac{n}{m}\right| \le \frac{1}{\sqrt{5} m^2}$$

Among all these pairs, at most finitely many of them satisfy $\left|\pi - \frac{n}{m}\right| \ge \pi\varepsilon$. After removing this finitely many exceptions, for the rest of $(m,n)$ pairs, we have

$$\left| \pi m - n \right| \le \frac{1}{\sqrt{5} m} = \frac{\pi}{\sqrt{5}n}\left[ 1 + \frac{1}{\pi}\left(\frac{n}{m} - \pi\right)\right] < \frac{\pi}{\sqrt{5}n} (1 + \varepsilon)$$ Using the identity $|\sin x| \le |x|$, we find for the corresponding $n$:

$$\frac{1}{n\sin^2 n} \ge \frac{1}{n (\pi m - n)^2} \ge \frac{5n}{\pi^2(1+\varepsilon)^2}$$

This implies $\displaystyle (-1)^n \frac{1}{n\sin^2 n}$ doesn't converge to $0$ as $n \to \infty$. As a result, the series $(*1)$ diverges.

For this question, you can use the Divergence Test, which states that if $f(x)$ does not tend towards $0$ as $x$ goes to infinity, $\sum_{n=1}^{\infty}f(n)$ diverges. In this case, $f(x)$ does not have a limit as $x$ goes to infinity, so $\sum_{n=1}^{\infty}\frac{1}{sin^2(n)}$ diverges.

Just observe that

$$\sin^2(n) < n .$$

Hint: What is the behavior of $\sin^2(n)$ as $n \rightarrow \infty$?

Remember: If $\sum a_n$ converges, then $\lim(a_n) = 0$.

Let $S_n$ be the partial sum, then $S_n \geq n$. So the series diverges to $\infty$

Thank you for your answers. I edited my original post and still didn't manage to prove or disprove the statement.