I am trying to argue that $$ \sum_{n=1}^{\infty} \frac{\sin(n)}{n} $$ is divergent. It get that it must be divergent because $\sin(n)$ is bounded and there is an $n$ on the bottom. But I have to use one of the tests in Stewart's Calculus book and I can't figure it out. I can't use the Comparison Tests or the Integral Test because they require positive terms. I can't take absolute values, that would only show that it is not absolutely convergent (and so it might still be convergent). The Divergence Test also doesn't work.

I see from this question:

Evaluate $ \sum_{n=1}^{\infty} \frac{\sin \ n}{ n } $ using the fourier series

that the series is actually convergent, but using some math that I don't know anything about. My questions are

(1) Is this series really convergent?

(2) Can this series be handled using the tests in Stewart's Calculus book?


One may apply the Dirichlet test, noticing that

  • $\displaystyle \frac1{n+1} \le \frac1{n}$

  • $\displaystyle \lim_{n \rightarrow \infty}\frac1{n} = 0$

  • $\displaystyle \left|\sum^{N}_{n=1}\sin n\right|=\left|\text{Im}\sum^{N}_{n=1}e^{in}\right| \leq \left|e^i\frac{1-e^{iN}}{1-e^i}\right| \leq \frac{2}{|1-e^i|}<\infty,\qquad N\ge1,$

giving the convergence of the given series.


We have $\enspace\displaystyle\sum\limits_{k=1}^\infty\frac{e^{i2\pi xk}}{k}=-\ln(1-e^{i2\pi x})\enspace$ for $\enspace 0<x<1\enspace$ .

Hint: $\enspace\displaystyle\sum\limits_{k=1}^\infty \frac{(re^{i2\pi x})^k}{k}=-\ln(1-re^{i2\pi x})\enspace$ for $|r|<1\enspace$ and

$\hspace{1cm}\enspace\displaystyle -\ln(1-re^{i2\pi x})\to -\ln(1-e^{i2\pi x})$ for $r\to 1^-$ if $0<x<1$


$\enspace\displaystyle -\ln(1-e^{i2\pi x})=-\ln(1-e^{i2\pi (x+n)})= -\ln(e^{i\pi (x-\frac{1}{2}+n)}2\sin(\pi x))$ $\hspace{3cm}\enspace\displaystyle = -i\pi (x-\frac{1}{2}+n)-\ln(2\sin(\pi x))\enspace$ with $\enspace n\in\mathbb{Z}$

With $\enspace\displaystyle x=\frac{1}{2}\enspace$ follows $\enspace n=0$ .

With $\enspace\displaystyle x=\frac{1}{2\pi}\enspace$ one gets $\enspace\displaystyle\sum\limits_{k=1}^\infty\frac{e^{ik}}{k}=-i\pi (\frac{1}{2\pi}-\frac{1}{2})-\ln(2\sin(\frac{1}{2}))$

and therefore $\enspace\displaystyle \sum\limits_{k=1}^\infty\frac{\sin(k)}{k}=\frac{\pi-1}{2}$ .

Note: I found Stewart's Calculus book under the link https://assassinezmoi.files.wordpress.com/2012/03/calculus-6th-edition-james-stewart.pdf . Use Fourier to get the result which I have written. Then you can show, that your series is convergent.

Obviously the hint above (one can see e.g. the literature) is not enough. Therefore a bit more text.

First: The logarithm is a continuous function.

Second: Be $\enspace\displaystyle 0<x_0<x\leq\frac{1}{2}\enspace$ or $\enspace\displaystyle \frac{1}{2}\leq x<x_0<1$ . Then we have $\enspace\displaystyle |e^{i2\pi x}-1|>|e^{i2\pi x_0}-1|$ .

It follows $\enspace\displaystyle |\lim\limits_{r\to 1^-}\sum\limits_{k=1}^n \frac{1-r^k}{1-r}\frac{e^{-i2\pi x k}}{k}|=|\frac{1-e^{-i2\pi x n}}{e^{i2\pi x}-1}|\leq |\frac{2}{e^{i2\pi x_0}-1}|\enspace$ and this inequality is independend of $n$ .

We get

$\displaystyle |\sum\limits_{k=1}^\infty \frac{e^{-i2\pi x k}}{k}-\lim\limits_{r\to 1^-}\sum\limits_{k=1}^\infty \frac{r^k e^{-i2\pi x k}}{k}|=|\lim\limits_{r\to 1^-}(1-r)\sum\limits_{k=1}^\infty \frac{1-r^k}{1-r}\frac{e^{-i2\pi x k}}{k}|$

$\hspace{7cm}\displaystyle \leq |\frac{2}{e^{i2\pi x_0}-1}|\lim\limits_{r\to 1^-}(1-r)=0$

and therefore with $\enspace 0<x<1$

$\displaystyle \sum\limits_{k=1}^\infty \frac{e^{-i2\pi x k}}{k}=\lim\limits_{r\to 1^-} -\ln(1-re^{-i2\pi x})=-\ln(1-e^{-i2\pi x})$ .

With the conjugated complex series we get what I had written in the first line.

  • $\begingroup$ How do you relate $\sum_{k=1}^{\infty} \frac{e^{2\pi x k}}{k}$ to the limit $\lim_{r\uparrow 1} \sum_{k=1}^{\infty} \frac{e^{2\pi x k}}{k} r^k$? $\endgroup$ Apr 10 '17 at 13:14
  • $\begingroup$ @Sangchul Lee : My answer to your question is in the second part above. :-) $\endgroup$
    – user90369
    Apr 10 '17 at 13:52
  • $\begingroup$ Your answer still assumes the existence of $S = \sum_{k=1}^{\infty} e^{-2\pi i x k}/k$. For instance, the existence of the sum $\sum_{k=1}^{\infty} \frac{1-r^k}{1-r} e^{-2\pi i x k}/k$ for some $|r| < 1$ implies the existence of $S$ as well since $S$ can be recovered by a linear combination of two 'convergent' sums $$S=(1-r)\sum_{k=1}^{\infty}\frac{1-r^k}{1-r}\frac{e^{-2\pi i xk}}{k}+\sum_{k=1}^{\infty}\frac{e^{-2\pi i xk}}{k}r^k$$ Clearly you want to circumvent this circular argument by some work-around. $\endgroup$ Apr 10 '17 at 14:22
  • $\begingroup$ @Sangchul Lee : Of course I assume first. But the limitation of $\displaystyle\sum\limits_{k=1}^n \frac{1-r^k}{1-r}\frac{e^{-i2\pi xk}}{k}$ independend of $n$ and the formula $\displaystyle\sum\limits_{k=1}^\infty \frac{(re^{-i2\pi x})^k}{k}=-\ln(1-re^{-i2\pi x})$ leads to the formula for the fourier series $\displaystyle\sum\limits_{k=1}^\infty \frac{e^{-i2\pi xk}}{k}$ . The OP should study Fourier, that's the best. $\endgroup$
    – user90369
    Apr 10 '17 at 14:35
  • 1
    $\begingroup$ @Sangchul Lee : Yes, I agree. --- But for me it's not clear how the OP likes to have his proof. Now he has the choise. :-) $\endgroup$
    – user90369
    Apr 10 '17 at 14:49

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