Convergence of a series $\sum {\left( {1 - \frac{{\sin {a_n}}}{{{a_n}}}} \right)} $ Given ${a_n}$ a converging series $\sum {{a_n}}$, and also that for every $n$, $a_n\ne0$.
Does the series $\sum {\left( {1 - \frac{{\sin {a_n}}}{{{a_n}}}} \right)} $ converge?
For every example I've tried I see that it does, but I can't find a "general" conclusion for this.
 A: The Taylor series expansion of $1-\dfrac{\sin x}{x}$ is $\dfrac{x^2}{6}-\dfrac{x^4}{120}+\cdots$. 
So, when $a_n \approx 0$, we have that $1-\dfrac{\sin a_n}{a_n}$ behaves like $\dfrac{1}{6}a_n^2$. 
Based on this, consider the sequences $a_n = \dfrac{(-1)^n}{\sqrt{n}}$ and $a_n = \dfrac{1}{n^2}$. For both of these sequences, the series $\displaystyle\sum_{n = 1}^{\infty}a_n$ converges. What can you say about $\displaystyle\sum_{n = 1}^{\infty}\left(1-\dfrac{\sin a_n}{a_n}\right)$ for both of these sequences?
A: (The following is a tad long, aiming at giving intuition. Skip to the last lines if you only want a conclusion)
Since the original series is convergent, $a_n \xrightarrow[n\to\infty]{} 0$. But then, 
$$
\sin a_n  = a_n - \frac{a_n^3}{6} + o(a_n^3)
$$
and therefore
$$
1-\frac{\sin a_n}{a_n} = \frac{a_n^2}{6}+o(a_n^2)
$$
If $\sum a_n$ were absolutely convergent (or $a_n$ non-negative), that would be great -- then the above would ensure $\sum_n (1-\frac{\sin a_n}{a_n})$ is also convergent by comparison (can you argue why?). But this hints that, otherwise (only conditional convergence of $\sum_n a_n$, this may not be the case.
In particular, what if $a_n = \frac{(-1)^n}{\sqrt{n}}$ (where we choose that based on the Taylor expansion above: i.e., using the insight given by the fact that we get $a_n^2$)? Then the above shows that
$$
1-\frac{\sin a_n}{a_n} = \frac{1}{6n}+o\left(\frac{1}{n}\right)
$$
But the RHS is the general term of a divergent (non-negative) series. By comparison, the series with general term defined by the LHS must also be divergent.
