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So I've tried to show that the series is convergent with a help of Dirichlet's theroem $\sum_{n = 0 }^\infty \frac{\cos(\frac{2n\pi}{3})}{n+1}$, but it turned out that the following sum diverges $\sum_{n = 0 }^\infty \cos(\frac{2n\pi}{3}) $

$\sum_{n = 0 }^\infty \cos(\frac{2n\pi}{3}) = 1,-\frac{1}{2},-\frac{1}{2},1,...$ and so on.

I also checked the necessary conditon

$\lim_{n \to \infty} \frac{\cos(\frac{2n\pi}{3})}{n+1}= 0 $

After showing the convergence of this series I should calculate the sum with a help of the following power series (that was also a hint to this exercise)

$\sum_{n = 0 }^\infty \frac{\cos(\frac{2n\pi}{3})}{n+1}x^{n+1}$

this part seems to be relatively easy but to take the derivative of the sum I ought to know the radius of the power series and that comes down to calculating (probably) another limit...

$\lim_{n \to \infty} \frac{\frac{\cos(\frac{2(n+1)\pi}{3})}{n+1}}{\frac{\cos(\frac{2n\pi}{3})}{n+2}}= \lim_{n \to \infty} (\frac{n+2}{n+1})\frac{\cos(\frac{2(n+1)\pi}{3})}{\cos(\frac{2n\pi}{3})} = ...$

user
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Bartek Zamojtel
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  • Don't use multi-line Latex for your titles – Jose Avilez May 06 '24 at 15:15
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    Eventhough the cosine series diverges, it is bounded. Dirichlet's Test only uses the fact that it is bounded and does not care if the series converges or diverges. – conan May 06 '24 at 15:28
  • Yes, use Dirichlet's test. Showing the partial sums of cosine are bounded will be much easier if you know that cosine is the real part of complex exponential. – Mark May 06 '24 at 15:35
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    In such a simple case, all this seems overkill. $\frac{\cos\frac{2n\pi}3}{n+1}\to0$ and $$\begin{align}\sum_{n=0}^\infty\frac{\cos\frac{2n\pi}3}{n+1}&=\sum_{k=0}^\infty\left(\frac1{3k+1}-\frac{1/2}{3k+2}-\frac{1/2}{3k+3}\right)\&=\frac12\sum_{k=0}^\infty\frac{9k+7}{(3k+1)(3k+2)(3k+3)}<\infty. \end{align}$$ – Anne Bauval May 06 '24 at 16:27
  • @AnneBauval This was my first idea too but Mark observed that in this way we were considering a partial sum for an alternating series and therefore the method was wrong. – user May 06 '24 at 16:39
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    @user Actually, I guess it can be turned into a correct proof, but you need to explain why the other subsequences tend to the same limit. It indeed follows from the general term tending to $0$, which Anne Bauval mentioned, but it has to be verified. In a calculus exam, I wouldn't accept such a solution without a detailed proof. – Mark May 06 '24 at 16:46
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    @Mark Thanks for this detailed explanation and for the previous comment. – user May 06 '24 at 16:48

2 Answers2

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We have that

$$\frac{\cos\left(\frac{2n\pi}{3}\right)}{n+1}= \Re\left(\frac{e^{i\frac{2n\pi}{3}}}{n+1}\right)$$

then since by geometric series

$$\sum_{n = 0 }^\infty \frac{x^n}{n+1}=-\frac{1}{x}\log(1-x)$$

we have

$$\sum_{n = 0 }^\infty \frac{\cos(\frac{2n\pi}{3})}{n+1}=\Re\left( \sum_{n = 0 }^\infty \frac{e^{i\frac{2n\pi}{3}}}{n+1}\right)=\Re\left(-e^{-i\frac{2\pi}{3}}\log\left(1-e^{i\frac{2\pi}{3}}\right)\right)=\frac{\log 3}4+\frac{\sqrt 3 \pi}{12}$$

Bob Dobbs
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user
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You can use Dirichlet's test. In order to apply it, you need to show that $\sum_{k=0}^{n-1} \cos\left(\frac{2k\pi}{3} \right)$ is less than some fixed $M\in \mathbb{R}$ for all $n$. This is easy since, as you mention, the series has a simple pattern. In fact, one can even show that $\sum_{k=0}^{n-1} \cos\left(a+k\cdot d \right)$ is bounded. This is not too hard to do thanks to the following formula: $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr).$$ You can find a proof of the formula here.

Vivaan Daga
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