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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?

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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.

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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 $0<x<1$

$\hspace{1cm}$

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

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  • $\begingroup$ How do you relate $\sum_{k=1}^{\infty} \frac{e^{2\pi x k}}{k}$ to the limit $\lim_{r\uparrow 1} \sum_{k=1}^{\infty} \frac{e^{2\pi x k}}{k} r^k$? $\endgroup$ – Sangchul Lee 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. $\endgroup$ – Sangchul Lee 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
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    $\begingroup$ @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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