# Convergence of $\sum_{n=1}^\infty n^2 \sin(\frac{π}{2^n})$

This is my series: $$\sum_{n=1}^\infty n^2 \sin(\frac{π}{2^n})$$

WolframAlpha says it converges, but I have no idea how to get the answer. I have learned comparison test, ratio test, root test and integral test. I don't really know which one of those to use. So far the only decent option seems the regular comparison test.

$\lim \limits_{x \to \infty} \frac{a_n}{b_n} = c, c \ne 0, c \ne \infty$

I tried something taking an geometric series for $b_k$ (like $\frac{1}{2^n}$ ) to get: $\lim \limits_{x \to \infty} \frac{a_n}{b_n} = \lim \limits_{x \to \infty} n^2 \sin(\frac{π}{2^n}) 2^n = \lim \limits_{x \to \infty} n^2 \frac{\sin(\frac{π} {2^n})}{\frac{π} {2^n}} \frac{π} {2^n} 2^n = \lim \limits_{x \to \infty} n^2 \frac{\sin(\frac{π} {2^n})}{\frac{π} {2^n}} π$

But that still comes to infinity. If i use harmonic series to get rid of infinity (n^2), I can't get rid of the 0 from sinus and if I use geometric series it's vice-versa.

The following inequality holds: $$\sin x \le x\qquad(x\ge 0)$$ Then $$0\le \sin\left(\frac{\pi}{2^n}\right)\le \frac{\pi}{2^n}.$$
• Using the inequality, we know $$0\le n^2\sin\left(\frac{\pi}{2^n}\right)\le \frac{n^2\pi}{2^n}.$$ Then all you need is to show $\sum \frac{n^2}{2^n}$ converges and apply comparison test. Commented Feb 18, 2016 at 3:30
Hint using $\sin x \approx x$ for small $x\ll1$ Then you only have to show that $$\sum_nn^22^{-n}$$ converges
We have: $$\sum_{n\geq 1}\frac{n^2}{2^{kn}}=\frac{2^k(2^k+1)}{(2^k-1)^3} \tag{1}$$ hence:
$$\sum_{n\geq 1}n^2\sin\left(\frac{\pi}{2^n}\right) = \sum_{n\geq 1}\sum_{m\geq 0}\frac{n^2 (-1)^m \pi^{2m+1}}{(2m+1)! 2^{(2m+1)n}}=\sum_{m\geq 0}\frac{(-1)^m (2\pi)^{2m+1}(2\cdot 4^m+1)}{(2m+1)!(2\cdot 4^m-1)^3}\tag{2}$$ where the RHS is a very fast-converging series. By adding just the first four terms we get: $$\sum_{n\geq 1}n^2\sin\left(\frac{\pi}{2^n}\right)\approx \color{red}{17.85}.\tag{3}$$