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$$\sum_1^\infty \cos\left( \frac{\pi n} 3 \right) \left(n^{\frac 1 {\sqrt[6]{n+6}}}-1\right)$$

I tried to use Dirichlet, but unsuccessfully. Please, give me hints. Thanks a lot.

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  • $\begingroup$ This and this are related, there are a few tips there that could be helpful. $\endgroup$ – kingW3 May 22 '17 at 21:10
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Convergence is not absolute. Since $x>0\implies e^x>1+x $, we have $$n>1\implies n^{(n+6)^{-1/6}}-1=e^{(n+6)^{-1/6}\ln n}-1>(n+6)^{-1/6}\ln n.$$

Let $y_n$ be the $n$th term of the series. For $k\in \mathbb N$ we have $\cos \pi (6k+1)/3=1/2,$ so we have $$y_{(6k+1)}>(1/2)(6k+7)^{-1/6}\ln (6k+1)>(1/2)(6^6k^6)^{-1/6}\ln (6k)=(1/2)(\ln 6k)/(6k). $$

$$\text {Therefore }\quad \sum_{n=1}^{\infty}|y_n|\geq \sum_{k=1}^{\infty}|y_{(6k+1)}|\geq (1/2)\sum_{k=1}^{\infty} (\ln 6k)/6k\geq (1/12)\sum_{k=1}^{\infty}1/k=\infty.$$

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  • $\begingroup$ This is a bit long-winded. Why not just say $ n^{(n+6)^{-1/6}}-1 >(n+6)^{-1/6} > 1/(n+6)$ for large $n$ and sum over $n=6,12,18,\dots?$ $\endgroup$ – zhw. May 23 '17 at 18:45
  • $\begingroup$ @zhw. Agreed. it's easy to get wrapped up in details. $\endgroup$ – DanielWainfleet May 23 '17 at 20:40
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\begin{align} & \Big( \cos \frac{1\pi} 3, \quad \cos\frac{2\pi} 3, \quad \cos\frac{3\pi} 3, \quad \cos\frac{4\pi} 3, \quad \cos \frac{5\pi} 3, \quad \cos\frac{6\pi} 3, \quad \ldots \Big) \\[10pt] = \frac 1 2 & \Big( \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \ldots\ldots\ldots \Big) \end{align} The sequence $1,1,0,-1,-1,0$ repeats forever.

So you have a sum of two positive terms, then a sum of two negative terms, etc.

The alternating series test says that if the terms decrease and approach $0$ and have alternating signs, then the series converges.

Whether it converges absolutely takes more work to answer after that. Maybe I'll be back.

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  • $\begingroup$ Convergence is not absolute. I have entered a proof of it. $\endgroup$ – DanielWainfleet May 23 '17 at 5:46
  • $\begingroup$ $\cos \pi /3=1/2.$ ... $(\cos \pi n/3)_n=(1/2,-1/2, -1, -1/2, 1/2,1)_{repeated}.$ ... I rarely vote down but I did. With the $n$th term being $ y_n$ and with $z_n=(1/2)y_{(3n-2)} -(1/2)y_{(3n-1)}-y_{(3n)} ,$ you can show that $z_n\to 0$ and that $(z_n)_n$ is an "alternating-decreasing" series for all sufficiently large $n$. $\endgroup$ – DanielWainfleet May 23 '17 at 5:59
  • $\begingroup$ This is not an alternating series. $\endgroup$ – zhw. May 23 '17 at 18:31
  • $\begingroup$ @zhw. : Let the series be $a_1+a_2+a_3+\cdots$ and let $$\begin{align} b_1 & = a_1+a_2 \\ b_2 & = a_4 + a_5 \\ b_3 & = a_7 + a_8 \\ b_4 & = a_{10} + a_{11} \\ & \,\,\, \vdots \end{align} $$ Then $b_1+b_2+b_3+\cdots$ is an alternating series, to which the alternating series test applies. (And the omitted $a_k$ when $k$ is a multiple of $3$ are all zero.) In general that might not mean the original series converges, but as applied in this particular instance, that does follow. $\qquad$ $\endgroup$ – Michael Hardy May 23 '17 at 18:35
  • $\begingroup$ @DanielWainfleet : I've corrected that clumsy oversight. $\endgroup$ – Michael Hardy May 23 '17 at 18:37

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