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How can I show this series converge without using Abel test(Dirichlet test) or integral test?

$\sum _{n=1}^{\infty }\:\frac{\sin\left(n\right)}{n}$

I have no idea to solve this only by using thms and properties related to convergence of sequences and subsequences.

leeway00
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  • You need to use that $\sum_{m=1}^n \sin(m)=\Im(\frac{1-e^{i(n+1)}}{1-e^i})$ is bounded ie. do a partial summation (which is how Dirichlet test is proven) – reuns Mar 31 '21 at 10:51
  • @ClaudeLeibovici Yes you are right. My mistake – leeway00 Mar 31 '21 at 10:59
  • An alternative: $\sum_{m=1}^n \frac{e^{im}}{m} = \int_0^1 x e^i\frac{1-x^n e^{in x}}{1-x e^i}dx$ and it is easy to see that $\lim_{n\to \infty} \int_0^1 |x e^i\frac{x^n e^{in x}}{1-x e^i}|dx=0$ – reuns Mar 31 '21 at 11:02

1 Answers1

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The convergence is generally established using Dirichlet's test and the value determined using $\sum \frac{z^n}n=-\ln(1-z)$.

It is possible however to establish convergence directly with the trick below, but we still need to go through the complexes.

Also since convergence is not granted we also have to work with partial sums.

Let's call $\quad S(N)=\sum\limits_{n=1}^N \dfrac{e^{in}}{n}\quad$ and multiply it by $(1-e^i)\neq 0$

$\begin{align}(1-e^i)S(N) &=\sum\limits_{n=1}^N \dfrac{e^{in}}{n}-\sum\limits_{n=1}^N \dfrac{e^{i(n+1)}}{n}\\ &=\sum\limits_{n=1}^N \dfrac{e^{in}}{n}-\sum\limits_{n=2}^{N+1} \dfrac{e^{in}}{n-1}&\text{shifting indexes}\\ &=\underbrace{\dfrac{e^{i1}}1}_\text{constant}-\underbrace{\dfrac{e^{i(N+1)}}{N}}_{\to 0}-\underbrace{\sum\limits_{n=2}^N \dfrac{e^{in}}{n(n-1)}}_\text{absolutely CV }&\text{using }\tfrac 1n-\tfrac 1{n-1}=\tfrac {-1}{n(n-1)}\\ \end{align}$

The last series is absolutely convergent by comparison with $\sim \sum\frac 1{n^2}$ and the remainder $\dfrac{e^{i(N+1)}}{N}\to 0$, therefore we can conclude that $S(N)$ itself is convergent.

zwim
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