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?

  • $\begingroup$ Use Dirichlet's criterion for $\sum \frac{\lvert x\rvert^n}{n}\cdot \sin \bigl(n(x+\pi)\bigr)$ for $x \in [-1,0]$. $\endgroup$ Commented Jan 4, 2018 at 11:54

1 Answer 1


$$\sum_{n\geq 1}\frac{x^n \sin(nx)}{n}$$ 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 $$ \left|\sum_{n\geq N}\frac{x^n \sin(nx)}{n}\right| = O\left(\frac{1}{N}\right). $$

  • $\begingroup$ @qbert: total convergence $\to$ uniform convergence is not flawed, but you may have uniform convergence over any compact subset of $(-1,1)$ by such argument, and still fail to have uniform convergence over $(-1,1)$ or $[-1,1]$. So at some point we need to exploit some structural property of the functions we are summing (continuity or harmonicity), or just bounding the tails $\sum_{n\geq N} f_n(x)$. $\endgroup$ Commented Jan 4, 2018 at 18:17
  • $\begingroup$ @Math_QED: I have completely re-designed my proof. Now the key steps are just summation by parts / Dirichlet's test and a lemma about $$\sup_{x\in[0,1]} x^n(1-x).$$ $\endgroup$ Commented Jan 4, 2018 at 18:57
  • $\begingroup$ Thanks for the answer! I don't have too much time now (I have exams to study), but I will look at it as soon as these are done! $\endgroup$
    – user370967
    Commented Jan 4, 2018 at 19:11

You must log in to answer this question.