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.

  • 6
    $\begingroup$ I think he was asking about how to obtain the value of the sum, not showing that it is convergent $\endgroup$
    – Basel J.
    Jul 8 '20 at 17:39

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