Convergence or Divergence of $\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$

Question

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?

Answer 1

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 $\sum \frac 1 {n^2}$.