How to prove that the series $\sum\limits_{n=1}^\infty {\sin^2n} $ diverges

Question

I want to use a divergence test to prove that $\lim_{n\to \infty} \sin^2n$ does not converge. So $\sum_{i=1}^\infty \sin^2 n $ diverge. But because $\pi$ is an irrational number. So I cannot use subsequence with $n=n\pi$ and $n=\frac{(2n-1)\pi}{2}$. So how can I prove that the sum diverges?

Where $n \in \Bbb{N}$

Beacuse n must be an integer. And the only way that I can make sure that I will land on the same point of a unit circle is using $\pi$. But because $\pi$ is an irrational number. So there will never be a sufficient n that reach $\pi$. Hence I will not land on the same place on a unit circle. — Wakeme UpNow, Jun 21 '16 at 15:05

score 6 · Answer 1 · edited Jun 21 '16 at 15:36

Lemma. If $\lvert \sin x\rvert<1/4$, then $\lvert \sin (x+1)\rvert>1/4$.

Proof. Observe that $\frac{\pi}{4}<1<\frac{\pi}{3}$, and hence $\sin 1>\frac{\sqrt{2}}{2}$. Also, $\lvert \cos x\rvert>\sqrt{1-\left(\frac{1}{4}\right)^2}=\sqrt{\frac{15}{16}}$. Hence $$ \lvert \sin(x+1)\rvert\ge \lvert \sin 1\cos x\rvert-\lvert \sin x\cos 1\rvert \\ \ge \frac{\sqrt{2}}{2}\frac{\sqrt{15}}{4}-\frac{1}{4}=\frac{\sqrt{30}}{8}-\frac{1}{4}>\frac{1}{2}-\frac{1}{4}=\frac{1}{4} . $$ This implies that $\{\sin n\}$ DOES NOT converge to zero, since if $\sin n\to 0$, then eventually $\lvert \sin n\rvert<1/4$, for all $n\ge n_0$, for some $n_0\in\mathbb N$. But the Lemma above proves that this is impossible!

Now that means $\sin n\not\to 0$, then also $\sin^2 n\not\to 0$, and hence the series $\sum \sin^2 n$ cannot converge.

score 1 · Accepted Answer · answered Jun 21 '16 at 15:13

It's something like you suggested. Consider arbitrary $n$. Then if $|sin(n)|<{1\over\sqrt{2}}$ then $n \in (\pi k -{\pi \over 4}, \pi k + {\pi \over 4})$ for some $k$. Hence $n+1$ or $n+2$ lying in interval $(\pi k + {\pi \over 4}, \pi k + {3\pi \over 4})$ and $max\{|\sin(n+1)|,|\sin(n+2)|\} \ge {1 \over \sqrt{2}}$. We just showed that for every $n$ $\max\{\sin^2(n),\sin^2(n+1),\sin^2(n+2)\}\ge{1\over2}$.

So there is sequence $\{n_k\}$ such that $\sin^2(n_k) \ge {1\over2}$ and $n_k \to \infty$. Obviously $\sum\limits_{n=1}^{\infty} sin^2(n) \ge \sum\limits_{k=1}^{\infty} sin^2(n_k) \ge \sum\limits_{n=1}^{\infty} {1\over2} = \infty$

score 1 · Answer 3 · answered Jun 21 '16 at 15:41

Consider the following integer sequence: $$ m_k = \left\lfloor\frac{\pi}{2}\right\rfloor +\left\lfloor2\pi\cdot 10^k\right\rfloor. $$ One may see that as $$ \lim_{k\to\infty}\left(2\pi\cdot 10^k - \left\lfloor2\pi\cdot 10^k\right\rfloor\right) = 0$$ we get that $$ \lim_{k\to\infty}\left(\frac{\pi}{2} + 2\pi\cdot 10^k - m_k\right) = \frac{\pi}{2} - \left\lfloor\frac{\pi}{2}\right\rfloor $$ which follows $$ \lim_{k\to\infty}\sin\left(\frac{\pi}{2} + 2\pi\cdot 10^k - m_k\right) = \sin\left(\frac{\pi}{2} - \left\lfloor\frac{\pi}{2}\right\rfloor\right) $$ or $$ \lim_{k\to\infty}\cos m_k = \cos\left\lfloor\frac{\pi}{2}\right\rfloor \Rightarrow \lim_{k\to\infty}\sin m_k = \sin\left\lfloor\frac{\pi}{2}\right\rfloor > 0. $$ Thus subsequence $\sin(m_k)$ converges to nonzero number and so does subsequence $\sin^2(m_k)$ so the whole sum diverges.

score 0 · Answer 4 · edited Apr 13 '17 at 12:21

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If $\{\sin^2n\}$ converges to $0$, then so does $\{\sin n\}$. But $\{\sin n\}$ diverges as shown in this MSE question.

edited Apr 13 '17 at 12:21

Community

1

answered Jun 21 '16 at 15:07

Spenser

19,469

How is that? I don't see because ${\sin n}$ diverge ${{\sin n}^2}$ must diverge too. – Wakeme UpNow Jun 21 '16 at 15:14
Go to the definition. If there is a subsequence ${\sin n_k}$ such that $|\sin n_k|\geq \epsilon_0>0$ for all $k\geq 0$, then also $|\sin^2n_k|\geq\epsilon_0^2>0$ for all $k\geq 0$. Hence ${\sin^2n}$ diverges. – Spenser Jun 21 '16 at 15:16

How to prove that the series $\sum\limits_{n=1}^\infty {\sin^2n} $ diverges

4 Answers4