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Does $\sum_{n=1}^{\infty}\left(\frac{\sin x}{x}\right)^n$ converge for all values of x, except for $x=0$? I tried some values and it seems to make sense. Any proofs for this?

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    $\begingroup$ This is a geometric series with a common ratio of $\frac{\sin(x)}{x}$. Since $\left|\frac{\sin(x)}{x}\right|<1$ for all nonzero $x$, it converges for all $x\neq 0$ $\endgroup$ Mar 10 '21 at 21:21
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Note that $-1<\frac{\sin x}{x}<1$

let $u=\frac{\sin x}{x}$ then,

$$\begin{align}\sum_{n=1}^{\infty}\left(\frac{\sin x}{x}\right)^n&=\sum_{n=1}^{\infty}(u)^n\\ &=\lim \limits_{n \to \infty}\sum_{i=1}^{n}(u)^i\\ &=\lim \limits_{n \to \infty}\frac{1-u^{n+1}}{1-u}\\ &=\frac{\lim \limits_{n \to \infty}(1-u^{n+1})}{\lim \limits_{n \to \infty}(1-u)}\\&=\frac{1}{1-u}\end{align}$$

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