# Infinite sum of $\sum_{n=1}^\infty \sin^{2n}\left(\frac{\pi}{n}\right)$

I have tried looking online for methods for solving sums such as this one but i was only able to find how to solve sums that look like $$\sum_{n=1}^{\infty}\sin^{2n}\left(\frac{\pi}{a}\right)$$ or $$\sum_{n=1}^{\infty}\sin^{2}\left(\frac{\pi}{n}\right)$$ but not much on things like this. Calculator websites like wolframalpha give a value ($$1.49$$ ($$2$$ d.p)) but dont actually proof how to get this value.

$$\sum_{n=1}^{\infty}\sin^{2n}\left(\frac{\pi}{n}\right)$$

We have the trivial bound $$|\sin(x)| \leq |x|$$ for all $$x \in \mathbb{R}$$, so we can upper bound this series (which has nonnegative terms because the powers are even) as follows:

$$\sum_{n=1}^\infty \sin^{2k}(\frac{\pi}{n}) \leq \sum_{n=1}^\infty \frac{\pi^{2k}}{n^{2k}} = \pi^{2k} \sum_{n=1}^\infty \frac{1}{n^{2k}},$$

and this last is a p-series, hence convergent for $$k \geq 1$$.

The same method works if $$k = n$$ to show that

$$\sum_{n=1}^\infty \sin^{2n}(\frac{\pi}{n}) \leq \sum_{n=1}^\infty \frac{\pi^{2n}}{n^{2n}} = \sum_{n=1}^\infty \left( \frac{\pi}{n} \right)^{2n},$$

and the nth root test shows that this latter series converges.

• I think he was asking about how to obtain the value of the sum, not showing that it is convergent Jul 8 '20 at 17:39