Uniform Convergence of Series $\sum_{n=1}^{\infty}\frac{\sin(x)^n}{n}$

Question

I'm trying to show uniform convergence of a series of complex numbers, but I'm having trouble. The series is as follows: $$\sum_{n=1}^{\infty}\frac{\sin(x)^n}{n} \rm{~~~~~~for}~~~0<x<\pi/2$$ I honestly have no idea where to start with this. I know the top will get very small very quickly... does that show convergence?

Answer 1

To formalize your thought that the top gets small quickly, we could notice that the ratio of consecutive terms is: $$\frac{\frac{\sin(x)^{n+1}}{n+1}}{\frac{\sin(x)^n}{n}}=\frac{n+1}{n}\sin(x)$$ which tends to $\sin(x)$ as $n$ goes to infinity - so, this, in the sense of convergence, acts kind of like a geometric series and hence converges when $\sin(x)<1$. This is probably essentially what you noticed.

However, the series diverges when $\sin(x)=1$, since it is, in that case, equal to the harmonic series. This is problematic, because we can see that, even though $x=\frac{\pi}2$, where $\sin(x)=1$, is outside the domain, it is on the boundary of the domain and as we approach that boundary, $\sin(x)$ will approach $1$ and the series will, termwise, approach the harmonic series, and hence the sum will increase without bound and cannot thus be uniformly convergent.

(To be very formal, for any $n$, we can choose $x\in(0,\pi/2)$ such that $\sin(x)^k>\frac{1}2$ for all $k\leq n$, and then your sum is at least $\frac{1}2\sum_{i=1}^{n}\frac{1}i$, where the last sum can be forced to be arbitrarily large)

Answer 2

As pointed in the other answer, the series is pointwise convergent on $\left(0,\frac{\pi}{2}\right)$ by Dirichlet's test, since: $$\sum_{n=1}^{N}\sin(x)^n \leq \frac{\sin x}{1-\sin x}$$ and $\frac{1}{n}$ decreases towards zero. However, in a left neighbourhood of $x=\frac{\pi}{2}$ the original series behaves like the harmonic series: $$ \sum_{n=1}^{N}\frac{\sin(\pi/2-\varepsilon)^n}{n}=\sum_{n=1}^{N}\frac{\cos(\varepsilon)^n}{n}\geq\sum_{n=1}^{N}\frac{1}{ne^{n\varepsilon^2}}\geq e^{-N\varepsilon^2} H_N$$ hence it is sufficient to take $\varepsilon\approx\frac{1}{\sqrt{N}}$ to prove that we cannot have uniform convergence.