Suppose $\sum x_n$ converges (not necessarily absolutely), does $\sum \sin x_n$ necessarily converge? The question is:

If $\sum x_n$ converges, does $\sum \sin x_n$ converge?

I know that if $\sum x_n$ converges absolutely, then $\sum \sin x_n$ converges. My intuition is that we cannot completely abandon absolute convergence, but I am struggling to come up with a counterexample.
 A: You are correct, without absolute convergence this is not necessarily true. Here is a construction of a counterexample, though not a very elegant one: Let $a_n = n^{-1/3}$ for $n=1,2,\ldots$, so that $\sum_{n=1}^\infty (a_n - \sin a_n) \approx \sum_{n=1}^\infty a_n^3/6 = \sum_{n=1}^\infty n^{-1}/6 = \infty$ diverges. Define $y_{2k-1} = y_{2k} =a_k$, so that the alternating series $\sum_{n=1}^\infty (-1)^{n-1} y_n = a_1 -a_1 + a_2 -a_2 \pm \ldots$ obviously converges to zero. Now replace every positive term $y_n$ in this series with $2^{n}$ terms $x_k = 2^{-n} y_n$, i.e., pass to the series $a_1/2+a_1/2-a_1 + a_2/4+a_2/4+a_2/4+a_2/4-a_2 \pm \ldots$. Obviously this series still converges to zero. However, the series $\sum_{n=1}^\infty \sin x_n$ will diverge. The reason for this is that the series of differences $\sum_{n=1}^\infty (x_n - \sin x_n)$ diverges, and the reason for that is that the positive terms of this series are $\sum_{n=1}^\infty 2^n(2^{-n}a_n)^3 \approx \sum_{n=1}^\infty 4^{-n}n^{-1}/6 < \infty$, whereas the negative terms are $\sum_{n=1}^\infty (-a_n+\sin a_n) \approx \sum_{n=1}^\infty (-n^{-1}/6) = -\infty$.
A: Let's define a sequence $a_n$ three terms at a time. The $n$th grouping will look like $$\frac{1}{n^{1/3}}, \frac{-1}{2 n^{1/3}}, \frac{-1}{2n^{1/3}}.$$
Then $\sum a_n$ converges (in fact the sum is $0$). Now recall $\sin x = x - x^3/6+O(x^5).$ The $n$th grouping for $\sin a_n$ then looks like 
$$\frac{1}{n^{1/3}}-\frac{1}{6n} + O(1/n^{5/3}),\frac{-1}{2n^{1/3}} + \frac{1}{48n} + O(1/n^{5/3}), \frac{-1}{2n^{1/3}} + \frac{1}{48n} + O(1/n^{5/3}).$$
The sum of those terms is $-1/(8n) + O(1/n^{5/3}).$ Therefore $\sum \sin a_n $ diverges to $-\infty.$
