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I'm trying to prove that $\sum_{n=1}^{\infty} (-1)^n \sin(\frac{1}{n})$ is convergent. I know that the limit to $\sin(\frac{1}{n})$ is $0$, and now I need to prove that $\sin(\frac{1}{n})$ is a decreasing function. How do I show that $\sin(\frac{1}{n})$ is decreasing?

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    $\begingroup$ You could prove that $\frac{d}{dx} \sin(1/x)$ is negative on the interval $(1, \infty)$. $\endgroup$ – Hyperion Feb 11 at 5:24
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Using the Prosthaphaeresis Reverse Identity

$$\sin(x)-\sin(y)=2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$

with $x=\frac1{n+1}$ and $y=\frac1n$ reveals for $n\ge 1$

$$\sin\left(\frac1{n+1}\right)-\sin\left(\frac1n\right)=-2\sin\left(\frac{1}{2n(n+1)}\right)\cos\left(\frac{2n+1}{2n(n+1)}\right)<0$$

And we are done!

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The derivative is

$$ - (1/n^2) \cos (1/n) $$

As long as $n>1$ the first factor is always positive and so is the second one so the entire thing is negative.

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You don't need to prove that $\sin 1/n$ is decreasing: $$\sin\frac1n=\frac1n+a_n$$ where $a_n=O(n^{-3})$, so that $$\sum(-1)^n\sin\frac1n=\sum(-1)^n\frac1n+\sum(-1)^na_n.$$ The former sum is convergent, by Leibniz, and the latter sum is absolutely convergent

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Observe that $\frac 1 n \in \left (0, \frac {\pi} {2} \right ) ,\ \text {for all}\ n \in \Bbb N.$ Also note that $\sin x$ is strictly increasing in $\left (0, \frac {\pi} {2} \right ).$ Here $0 < \frac {1} {n+1} < \frac 1 n < \frac {\pi} {2}$ and hence $\sin \left (\frac {1} {n+1} \right) < \sin \left (\frac 1 n \right),\ \text{for all}\ n \in \Bbb N.$ This proves that the sequence $\left \{\sin \left (\frac 1 n \right ) \right \}$ is strictly decreasing.

QED

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