<s>[INST] Convergence or Divergence of $\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$ 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? 
 [/INST] 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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