If $\sum_{n=1}^{\infty} a_{n}$ converges, then$\sum_{n=1}^{\infty} \sin(a_{n})$ also converges

This is a "Community Wiki" answer recording a comment by Daniel Fischer under the question. The comment provides a link The set of functions which map convergent series to convergent series to a proof that the result in question is false in general, though certainly true when the $a_n$ are non-negative. My reason for writing this answer is that comments can vanish more easily than answers and can probably also be more easily overlooked.

If $(a_n)_n$ is assumed non-negative (or non-positive), this follows from the comparison theorem as $0 \leq |\sin a_n| \leq a_n$.

However, the conclusion is false otherwise. Here is a counterexample: we are given in this answer an explicit sequence $(a_n)_n$ such that

• $\sum_n a_n$ converges
• $\sum_n a_n^3$ diverges
• $\sum_n a_n^4$ converges (by inspection of the explicit sequence given)

Specifially: for all $n\geq 1$, $$ a_{3n-2} = \frac{1}{n^{1/3}}, \quad a_{3n-1} = a_{3n} = -\frac{1}{2n^{1/3}} $$

In particular, clearly, $\lim_{n\to\infty}a_n =0$. Since $\sin x = x-\frac{x^3}{6} + O(x^4)$, we get $$ \sum_n \sin(a_n) = \sum_n a_n -\frac{1}{6}\sum_n a_n^3 + O\left(\sum_n a_n^4\right) $$ (the use of $O(\cdot)$ here is OK, as we deal with an absolutely convergent series at that point). But $\sum_n a_n, \sum_n a_n^4$ are convergent (convergent and absolutely convergent, respectively), while $\sum_n a_n^3$ isn't: so the RHS diverges. So the LHS must diverge too.