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How to determine convergence of this series. $$\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$$ I've tried using comparison test:

$\sin(\pi/n) \leq \pi/n $, so:

$$(3^{1/n}-1)\sin(\pi/n)<(3^{1/n}-1)\pi/n < (3^{\frac{1}{n}})\frac{\pi}{n}$$ By comparison test if $\sum(3^{\frac{1}{n}})\frac{\pi}{n}$ is convergent, so would be initial.

But $\sum(3^{\frac{1}{n}})\frac{\pi}{n}$ is divergent.

I also know that $\sin(\pi/n)$ is divergent. How would it help? Can you give a hint?

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  • $\begingroup$ Well, $\int_1^{\infty}(3^{1/x}-1)\sin(\pi/x)dx$ converges. Thus the series does too. $\endgroup$ – clathratus Oct 26 '18 at 16:32
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HINT

We have that

  • $3^x = e^{x\log 3}=1+x\log 3 +O(x^2)$
  • $\sin x = x +O(x^2)$

therefore

$$(3^{1/n}-1)\sin(\pi/n)\sim \frac {\pi\log 3} {n^2}$$

then refer to limit comparison test with $\sum \frac 1 {n^2}$.

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HINT

Note that $3^{1/n} \to 1$, so $3^{1/n} - 1 \to 0$ and the last step in your inequality should not be made.

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Your series is absolutely convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{\log 3}{n}\cdot\frac{\pi}{n}=\frac{\pi^3}{6}\log(3)$.
If you like explicit bounds, you may notice that $$ \frac{3}{2} = \frac{2n+1}{2n}\cdot\frac{2n+2}{2n+1}\cdot\ldots\cdot\frac{2n+n}{3n-1} ,\qquad 2=\frac{n+1}{n}\cdot\frac{n+2}{n+1}\cdot\ldots\cdot\frac{n+n}{2n-1}$$ $$ 3 = \prod_{k=1}^{n}\frac{(2n+k)(n+k)}{(2n+k-1)(n+k-1)}=\prod_{k=1}^{n}\frac{(2n+k)(2n-k+1)}{(2n+k-1)(2n-k)} $$ and by the AM-GM inequality $$ 3^{1/n} \leq \frac{1}{n}\sum_{k=1}^{n}\frac{(2n+k)(2n-k+1)}{(2n+k-1)(2n-k)}=\frac{1}{n}\sum_{k=1}^{n}\frac{(4n+1)^2-(2k-1)^2}{(4n-1)^2-(2k-1)^2} $$ or $$ 3^{1/n}-1 \leq \sum_{k=1}^{n}\frac{16}{(4n-1)^2-(2k-1)^2}=\sum_{k=1}^{n}\frac{4}{(2n-k)(2n+k-1)}. $$ By letting $H_n = \sum_{k=1}^{n}\frac{1}{k}$ the previous line gives $$ 3^{1/n}-1 \leq \frac{4}{4n-1}\sum_{k=1}^{n}\left(\frac{1}{2n-k}+\frac{1}{2n+k-1}\right)=\frac{4}{4n-1}\left[H_{3n-1}-H_{n-1}\right] $$ and by the Hermite-Hadamard inequality it follows that $$ 3^{1/n}-1 \leq \frac{4}{4n-1}\left(\log 3+\frac{4}{3(4n-1)}\right)\leq \frac{1}{n}\left(\log(3)+\frac{1}{n}\right). $$ In particular the given series is bounded by $\frac{\pi^3}{6}\log(3)+\pi\zeta(3)$.

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  • $\begingroup$ Then you finally leave as a moderator! I'm really sorry for that but I'm happy to continue to read your answer. Ofetn I understand a 1% but they are very stimulating to learn more advanced topics and related methods. $\endgroup$ – user Oct 27 '18 at 7:57

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