Show uniform convergence of the series of functions $\sum_{n=1}^\infty
\frac{x^n \sin(nx)}{n}$ on the interval $[-1,1]$
My attempt: I showed the series converges uniformly on the interval $[-1/2,1/2]$ using Weierstrass M-test. I also showed the series converges uniformly on the interval $[1/2,1]$ by using Dirichlet's criterium, where I used that $\left|\sum_{k=1}^n \sin(kx)\right| \leq \frac{1}{\sin(x/2)}$
However, I'm stuck at showing it converges uniformly on the interval $[-1,-1/2]$. I tried to apply Dirichlet's criterium but can't conclude anything because of the behaviour of the term $x^n/n$ (which does not decrease monotonically).
Any ideas?
Answer
$$\sum_{n\geq 1}\frac{x^n \sin(nx)}{n}$$
is pointwise convergent for any $x\in[-1,1]$: that is trivial if $|x|<1$ and it follows from Dirichlet's test if $|x|=1$. In order to prove uniform convergence, it is enough to show that
$$ E(N) = \sup_{x\in[-1,1]}\left|\sum_{n\geq N}\frac{x^n \sin(nx)}{n}\right| $$
fulfills $\lim_{N\to +\infty}E(N)=0$. If $|x|<1$ we have
$$ \left|\sum_{n\geq N}\frac{x^n \sin(nx)}{n}\right|\leq \frac{1}{N}\sum_{n\geq N}|x|^n = \frac{|x|^N}{N(1-|x|)}.$$
For a fixed $N$, let
$ S_M(x) = \sum_{n=N}^{M}\sin(nx)$. We know that $|S_M(x)|\leq \frac{1}{|\sin(x/2)|}\leq\frac{\pi}{|x|}\leq 2\pi$ for any $x\in[-1,1]$ such that $|x|>\frac{1}{2}$, and by summation by parts
$$ \sum_{n\geq N}\frac{x^n\sin(nx)}{n} = \sum_{n\geq N}S_n(x)x^n\left(\frac{1}{n}-\frac{x}{n+1}\right). $$
If $x$ is negative the RHS can be bounded through the alternating series test, and it turns out to be $O\left(\frac{1}{N}\right)$. If $x$ is positive the RHS can be written as
$$\sum_{n\geq N}\frac{S_n(x)x^n}{n(n+1)}+\sum_{n\geq N}\frac{S_n(x)x^n(1-x)}{n+1} $$
where the first term is bounded by $\frac{2\pi}{N}$ in absolute value. We have
$$ \sup_{x\in[0,1]} x^n(1-x) \leq \frac{1}{en} $$
and this completes the proof that
$$ \left|\sum_{n\geq N}\frac{x^n \sin(nx)}{n}\right| = O\left(\frac{1}{N}\right). $$
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