How can I prove that if $a_n \neq 0$ for every $n$, then
$\sum _{n=1}^{\infty} \left(1- \frac{\sin(a_n)}{a_n}\right)$
converges if and only if
$\displaystyle{\sum_{n=1}^{\infty} a_n^2}$
converges?
How can I prove that if $a_n \neq 0$ for every $n$, then
$\sum _{n=1}^{\infty} \left(1- \frac{\sin(a_n)}{a_n}\right)$
converges if and only if
$\displaystyle{\sum_{n=1}^{\infty} a_n^2}$
converges?
Hint: $\begin{align} \sin(x) = &x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 + \dots. \\ &\Downarrow \\ \frac{\sin(x)}{x} = &1 - \frac{1}{3!}x^2 + \frac{1}{5!}x^4 + \dots \\ &\Downarrow \\ 1 - \frac{\sin(x)}{x} = &\frac{1}{3!}x^2 - \frac{1}{5!}x^4 + \dots \end{align}$
So for small $x$, we have $1 - \frac{\sin(x)}{x} = cx^2 + o(x^4)$. Let $x = a_n$, and note that $a_n \to 0$, necessarily.