TITLE: how to show that $\sin(\frac{1}{n})$ is a decreasing function
QUESTION [4 upvotes]: 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?

REPLY [3 votes]: 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

REPLY [2 votes]: 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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