How can I show the divergence of $\cos^3\frac{2n\pi}3$

Asked Apr 13 '18 at 02:00

Active Apr 13 '18 at 02:37

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How can I show the divergence of $$\sum_{n=1}^{\infty} \frac{1}{n} \cos^3 \frac{2n\pi}{3}?$$

I have noticed that the values of $\cos^3 \dfrac{2n\pi}{3}$ alternate between $\dfrac{1}{8}$ and $-\dfrac{1}{8}$.

I was hoping to show the partial sums are unbounded somehow, but since the values can be negative, I am unable to find a lower bound of each term in the sequence.

edited Apr 13 '18 at 02:37

Ѕᴀᴀᴅ

34,263

asked Apr 13 '18 at 02:00

Jungleshrimp

1,585

Use this https://en.wikipedia.org/wiki/Dirichlet%27s_test (are you sure it diverges?) – Reveillark Apr 13 '18 at 02:04
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You cannot. :-) – Przemysław Scherwentke Apr 13 '18 at 02:08
According to this post it does https://math.stackexchange.com/questions/2216555/general-infinite-series-convergence-or-divergence – Jungleshrimp Apr 13 '18 at 02:12
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HINT: $$\sum_{n=1}^{3N}\frac{\cos^3(2\pi n/3)}{n}=-\frac18\sum_{n=1}^N \left(\frac{1}{3n-2}+\frac{1}{3n-1}-\frac{8}{3n}\right)$$Now look at $\sum_{n=1}^{3N+1}\frac{\cos^3(2\pi n/3)}{n}$ and $\sum_{n=1}^{3N+2}\frac{\cos^3(2\pi n/3)}{n}$ – Mark Viola Apr 13 '18 at 02:14

How can I show the divergence of $\cos^3\frac{2n\pi}3$

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