how to show that $\sin(\frac{1}{n})$ is a decreasing function 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?
 A: 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!
A: 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.
A: 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
A: 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
