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

Using the limit comparison test twice I determined it converges however the process was longer than I thought necessary. I feel like I am overlooking a simpler solution, what other methods would be viable?

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    $\begingroup$ The denominator goes to $1$ and is therefore irrelevant. $\sin(\frac{1}{n^2})\leq \frac{1}{n^2}$, so the series converges. $\endgroup$ – Gabriel Romon Nov 1 '17 at 21:09
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Use equivalents;

$\sin\dfrac1{n^2}\sim_\infty\dfrac1{n^2}$, $1-\sin\frac1n\sim_\infty 1$, hence $$\frac{\sin\frac1{n^2}}{1-\sin\frac1n}\sim_\infty\frac1{n^2},\quad\text{which converges.}$$

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Hint: Can you think of a convenient upper bound on $\sin(1/n^2)$? Similarly, how does $1 - \sin(1/n)$ behave as $n \to \infty$?

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HINTS:

Note that we have

$$\sin(1/n^2)\le \frac1{n^2}$$

and for $n>1$

$$\frac{1}{1-\sin(1/n)}\le \frac1{1-1/n}$$

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