# 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?

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

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!

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.

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

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