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This question already has an answer here:

I have the series $\sum_{n=1}^\infty{\sin^2\left(\frac{1}{n}\right)}$ and I'm trying to examine whether it converges or not.

My Attempts:

  1. I first tried finding whether it diverges by checking if $\lim_{n\to\infty}{\sin^2\left(\frac{1}{n}\right)} \ne 0$.

$$ \lim_{n\to\infty}{\sin^2\left(\frac{1}{n}\right)}= \lim_{n\to\infty}{\sin\left(\frac{1}{n}\right)}\cdot\lim_{n\to\infty}{\sin\left(\frac{1}{n}\right)}=0 $$

  1. Since I didn't get a confirmation from the first try, I then tried the d'Alembert's Criterion which didn't get me very far.

$$ \frac{a_{n+1}}{a_n}= \frac{\sin^2\left(\frac{1}{n+1}\right)}{\sin^2\left(\frac{1}{n}\right)}= \frac{ -\dfrac{2\cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)}{n^2} }{ -\dfrac{2\cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n+1}\right)}{\left(n+1\right)^2} }= \frac{ \cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)\left(n+1\right)^2 }{ \cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n+1}\right)n^2 }=\ ... $$

  1. Finally, I tried Cauchy's Criterion, but I didn't get any conclusive result either.

$$ \sqrt[n]{a_n}= \sqrt[n]{\sin^2\left(\frac{1}{n}\right)}= \sin^{\frac{2}{n}}\left(\frac{1}{n}\right)=\ ... $$

Question:

I've been thinking for while of using the Comparison Test, but I'm not sure which series to compare mine to. How can I examine whether the series converges or not?

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marked as duplicate by Martin R, Ethan Bolker, Hans Engler, Guacho Perez, user99914 Mar 8 '18 at 22:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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HINT:

For all $x$, we have

$$0\le \sin^2(x)\le x^2$$

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Just use the fact that$$(\forall n\in\mathbb{N}):\sin^2\left(\frac1n\right)\leqslant\frac1{n^2}.$$

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Observe that $$ \sin^2(1/n)\sim\frac{1}{n^2}. $$ Indeed, $$ \lim_{n\to\infty}\frac{\sin^{2}(1/n)}{1/n^2}=\left[\lim_{n\to\infty}\frac{\sin(1/n)}{1/n}\right]^2=1 $$ where in the last equality we used the well known limit $$ \lim_{x\to 0}\frac{\sin x}{x}=1. $$ The limit comparison test implies that your series converges.

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As an alternative note that

$$\sin^2\left(\frac{1}{n}\right)\sim \frac1{n^2}$$

then use limit comparison test with $\sum_{n=1}^\infty\frac1{n^2}$.

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Using the help of the majority of the given answers suggesting that $\sin^2(\frac{1}{n})\le\frac{1}{n^2}$, I came up with the following two-step solution:

  1. Using the Direct Comparison Test:

    If the infinite series $\sum b_{n}$ converges and $0\leq a_{n}\leq b_{n}$ for all sufficiently large $n$ (that is, for all $n>N$ for some fixed value $N$), then the infinite series $\sum a_{n}$ also converges.

    Let $a_n=\sin^2\left(\frac1n\right)$ and $b_n=\frac{1}{n^2}$. Since $a_n \le b_n \ \ \forall\ \ n \in \mathbb{N}$, then, if the infinite series $\sum b_n$ is convergent $\sum a_n$ is as well.

  2. Using Riemann's Zeta Function:

    The following infinite series converges for all complex numbers s with real part greater than 1, and defines ζ(s) in this case: ${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }n^{-s}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots \ \ \ \ \ \sigma =\operatorname {Re} (s)>1.}$

    Since $ζ$ is convergent for $n>1$ and $\sum b_n=ζ(2)$, $\sum b_n$ is convergent, thus proving that $\sum a_n$ is convergent as well, based on step (1).

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