$\sum a_n$ converge but $\sum \sin(a_n)$ diverges? Problem :
Show there is $a_n$ such that
$\sum a_n$ converge but $\sum \sin(a_n)$ diverges
or show the $a_n$ doesn't exist.

My Attempt
Since $$|\sin x| \le |x|$$
$a_n$ changes its sign (If $a_n > 0$ or $a_n<0$ for all $n\in\mathbb{N}$, we know $\sum \sin (a_n)$ converge since $\sum a_n$ converge.)
I tried to use a local inverse of $f\colon x \to x - \frac{x^3}{6}$ at $x=0$ because $$\sin x \approx x - \frac{x^3}{6}$$
So I think there is $a_n = f^{-1}(x_n) \quad (x_n \to 0)$ such that
$$\sum a_n<\infty,\quad \sum \sin(a_n) \approx \sum x_n = \infty$$
But I can't construct it, or I was wrong.
So my question is I want to know an example of $a_n$ and how to construct it if $a_n$ exists.
If $a_n$ doesn't exist, I want to know how to prove it.
Thanks for help!
 A: The example itself
To expand on the great example that was presented in the comments, if we take
$a_n=b_nc_n$ where $b_n=\lfloor n/3 \rfloor^{-1/3}$ and $c_n=1,1,-2,1,1,-2,\ldots$, then since the partial sums of $\sum c_n$ are bounded and $b_n\to 0$ monotonically, then by Dirichlet's test the series $\sum a_n$ converges.
Now for $\sum \sin(a_n)$. Notice that every $3$ consecutive terms of $b_n$ which start with $n=3k$ are equal to one another. (So that in $a_n=b_nc_n$ the sum of every consecutive triplet starting with $n=3k$ is zero). Let's see what kind of cancellations we get for $\sin(a_n)$. If we sum each $3$ consecutive terms
$$\sin(k^{-1/3})+\sin(k^{-1/3})-\sin(2k^{-1/3})$$
using the Taylor series expansion $\sin(x)=x-\frac{x^3}{3!}+O(x^5)$, we get
$$
2\cdot(k^{-1/3}-k^{-1}/3!+O(k^{-5/3})) -(2k^{-1/3}-8k^{-1}/3!+O(k^{-5/3}))$$
which simplifies to $-\frac{10}{3!}\cdot \frac{1}{k} + O(k^{-5/3})$.
This means that the partial sum $\sum_{n=1}^m \sin(a_n)$ for $m$ a multiple of $3$ equals $$-\frac{10}{3!}\sum_{n=1}^{m/3} \frac{1}{n}+\sum_{n=1}^{m/3}O(n^{-5/3}),$$ so as $m\to\infty$ the right sum converges (since it is absolutely convergent) and the left sum diverges (since it is a constant times the harmonic series). Therefore $\sum_{n=1}^{\infty} \sin(a_n)$ diverges.

Motivation for constructing this example

*

*On first glance it seems reasonable to guess that it wouldn't be possible to construct such $a_n$, since $\sin x \sim x$ for $x\to 0$. This indeed shows that there can be no such $a_n$ with positive terms, as was mentioned in the question. However, when $a_n$ has alternating terms it is possible that we will get different kind of cancellations in $a_n$ and $\sin(a_n)$, because while $\sin(a_n)$ tends to $0$ "at the same rate" as $a_n$ does, these are of course not the exact same terms -- and so it is reasonable that we can construct a special $a_n$ such that the cancellations of $a_n$ and $\sin(a_n)$ would yield different results. This is my intuition for why it might be possible to construct such $a_n$.


*In order to try and get different cancellations for $\sin(a_n)$ we use the Taylor series of $\sin x$. The idea is to try and get the linear terms to cancel out, and the $x^3$ terms not to cancel each other and to diverge, and the $x^5$ terms (and above) to converge. So other than the constants that will guarantee the cancellation of the linear terms, let's try $a_n$ which is roughly of the form $n^{-\alpha}$ but such that $3$ consecutive terms are equal (see why $3$ below) so we take $\lfloor n/3 \rfloor^{-\alpha}$. We need $n^{-3\alpha}$ to diverge and $n^{-5\alpha}$ to converge, that is, we want $3\alpha\leq 1$ and $5\alpha>1$ so any $\frac{1}{5}< \alpha\leq\frac{1}{3}$ will do, and we choose $\alpha=\frac{1}{3}$. Now in order to get the linear terms to cancel out we multiply $\lfloor n/3\rfloor ^{-1/3}$ by the constants $1,1,-2,1,1,-2,\ldots$. (The simpler choice of $1,-1,1,-1,\ldots$ and $\lfloor n/2 \rfloor$ would not work, because that would make the $x^3$ terms, and all subsequent terms, cancel out each other as well. This is clear without resort to Taylor series, $\sin x$ is an odd function, so for $a_n=(-1)^n\lfloor n/2 \rfloor^{-\alpha}$ we get that the terms of $\sin(a_n)$ cancel each other in pairs just the same way as the terms of $a_n$ do).


*There is nothing special about $\sin(x)$, in the sense that for any non-linear $f(x)$ it is possible to find a convergent $\sum a_n$ such that $\sum f(a_n)$ diverges. See this question for details. I haven't read the answers yet but I believe the overall intuition should be the same-- that it is plausible that we can use non-linearity of $f(x)$ to get different cancellations for $\sum a_n$ and $\sum f(a_n)$.
