uniform convergence of $\sum_{n=1}^\infty (1-e^{-\frac xn})\sin(nx)$

Question

I have this question that I'm struggling to do in my calc textbook

Note that I have not been exposed to complex numbers yet

The question is: let $a$ be a positive real number such that $0 < a < \pi/2 $

Discuss whether the series of functions

$$\sum_{n=1}^\infty (1-e^{-\frac xn})\sin(nx)$$

converges pointwise on the closed interval $[a,\pi-a] $ and if so, whether the convergence is uniform

My effort:

I tried to prove that $(1-e^{-\frac xn})$ is convergent by itself because since $\sin(nx) \le 1$ but I couldn't figure it out as well.

I did some additional research on complex numbers and euler's formula but I couldn't really grasp it. Is it necessary to use euler's or is there something I'm missing out?

Thanks

Answer 1

For $0\le t\le 1$ we have $$0\le t-(1-e^{-t})=\sum_{n=2}^\infty (-1)^{n}{t^n\over n!}\le (e-2)\,t^2\le t^2$$ Therefore $$0\le {x\over n}-\left (1-e^{-{x\over n}}\right )\le {x^2\over n^2},\qquad n>\pi $$ Hence $$\left |{x\over n}\sin nx-\left (1-e^{-{x\over n}}\right )\sin nx\right |\le {x^2\over n^2},\quad n> \pi$$ The series $x^2\sum {1\over n^2}$ is uniformly convergent as the values of $x$ are bounded. Hence our series is uniformly convergent for $a\le x\le \pi-a,$ because the series $$x\sum_{n=1}^\infty {\sin nx\over x}$$ is uniformly convergent in that interval (by the Dirichlet test, see the accepted answer of series).

Answer 2

Here is a self-contained argument, only using basic trigonometry. Let $a_n=a_n(x)=1-e^{-x/n}$ and $A_N= A_N(x)=\sum_{n=1}^N a_n$. $\;$ Let $b_n=b_n(x)=\sin(nx)$ and $B_N= B_N(x)=\sum_{n=1}^N b_n$. Observe that $$2\cos(x/2)b_n =\sin(nx+x/2)-\sin(nx-x/2)\,,$$ so $$2\cos(x/2)B_N =\sin(Nx+x/2)-\sin(x/2) \,.$$ Since $\exists \gamma>0, \; \cos(x/2)\ge \gamma>0$ for $x \in [\alpha,\pi-\alpha]$, we deduce that $$B:=\sup_{x \in [\alpha,\pi-\alpha]} \sup_N |B_N(x)| <\infty \,.$$ Using summation in parts as in [1] gives that the partial sums $S_N =S_N(x)= \sum_{n=0}^N a_n b_n$ satisfy

$$S_M - S_N = a_M B_M - a_N B_N - \sum_{n=N}^{M-1} B_n (a_{n+1} - a_n)$$ for $N<M$. Thus, using monotonicity and positivity of $a_n$, $$|S_M - S_N| \le Ba_M+Ba_N+B(a_N-a_M) =2Ba_N\,.$$ Since $a_N(x) \to 0$ uniformly in $x \in [\alpha,\pi-\alpha]$, we conclude that $\{S_N(x)\}$ is a Cauchy sequence with respect to the supremum norm in $ [\alpha,\pi-\alpha]$, so it converges uniformly in that interval.

[1] https://en.wikipedia.org/wiki/Summation_by_parts#Applications