# 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?

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

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

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.

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 $$\frac{\pi^3}{6}\log(3)+\pi\zeta(3)$$.

• 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. – user Oct 27 '18 at 7:57