# convergence of sum of series

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?

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Source? Motivation? Failed approaches? – Did Jan 13 '12 at 15:50
Reading some calculus on my spare time, and found this "sentence" and tried proving it unsuccessfully. Motivation: Fun. Didn't manage to really start so no failed approaches. I only know that a_n goes to zero no matter which side we start from. – Guy Jan 13 '12 at 15:53
Then, (1) would you know how to prove that $a_n\to0$? And (2) assuming $a_n\to0$, would you be able to compare $1-\frac{\sin a_n}{a_n}$ with multiples of $a_n^2$? – Did Jan 13 '12 at 15:58
Are you familiar with the Maclaurin series for $\sin x$? – Brian M. Scott Jan 13 '12 at 16:02
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.