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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

Mittens
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jeff
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2 Answers2

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).

Ryszard Szwarc
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  • hey man thanks for the guidance, can I ask how does the first line comes about? Couldn't understand which theorem that was – jeff Apr 28 '22 at 15:56
  • I have used the series expansion of $e^{-t}$. Are you familiar with that ? If not, I can modify the proof, but then I need the MacLaurin formula of order $2.$ – Ryszard Szwarc Apr 28 '22 at 15:58
  • Ah no worries I did a search for it, just did not notice! No worries Sir I think this should be good enough, I will try to understand the rest. Thank you very much! – jeff Apr 28 '22 at 16:00
  • I have one more request if you don't mind, do you have any book to recommend for me to further understand these kinds of questions? – jeff Apr 28 '22 at 16:03
  • Any advanced calculus textbook should be fine. At the moment I cannot come up with any particular one. – Ryszard Szwarc Apr 28 '22 at 16:10
  • See https://math.stackexchange.com/questions/4040136/looking-for-advanced-single-variable-calculus-textbooks-not-real-analysis – Ryszard Szwarc Apr 28 '22 at 16:42
1

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

Yuval Peres
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