How to find $\lim_{n \to \infty}\Sigma_{i=1}^n\frac{\sin(\pi x)^2}{i^2}$?

Question

I am trying to find:

$$\lim_{n \to \infty}\sum_{i=1}^n\cfrac{\sin(\pi x)^2}{i^2}$$

using maple command:

limit(Sum(sin(Pi*x)^2/i^2, i = 1 .. n), n = infinity);

gives:

$\cfrac{1}{6} \left( \sin \left( \pi \,x \right) \right) ^{2}{\pi }^{2}$

How to find $\lim_{n \to \infty}\Sigma_{i=1}^n\frac{\sin(\pi x)^2}{i^2}$?

Answer 1

If you have an $x$ in the sum with $\sin(\pi x)^2$ ,it will just get out of the Summation. Thus,

$$\lim_{n \to \infty}\Sigma_{i=1}^n\frac{\sin(\pi x)^2}{i^2}$$ $$=\sin(\pi x)^2\lim_{n \to \infty}\Sigma_{i=1}^n\frac{1}{i^2}$$ $$= \frac{π^2×\sin(\pi x)^2}{6}$$

If you mean $\sin(\pi \color{red}{i})^2$ instead, $\sin(\pi \color{red}{i})^2 = 0 \forall i \in N $. So $$\lim_{n \to \infty}\Sigma_{i=1}^n\frac{\sin(\pi i)^2}{i^2} = 0$$

Answer 2

Note that: $$\lim_{n\to\infty}\sum_{i=1}^n\frac{\sin^2(\pi x)}{i^2}=\sin^2(\pi x)\sum_{i=1}^\infty\frac{1}{i^2}=\frac{\pi^2}{6}\sin^2(\pi x)$$ since: $$\sum_{i=1}^\infty\frac{1}{i^2}=\zeta(2)=\frac{\pi^2}{6}$$ Where $\zeta$ is the zeta function