The sum is $$\sum_{k=1}^{\infty}\frac{1}{k}\left( \frac{\sin{k} + 2}{3} \right) ^k$$
Here's what I've tried so far:
Root test (in its stronger limsup form), gives nothing, so I didn't bother with the ratio test. This also rules out any hope of forming bounds and comparing to another (geometric) series.
Actual numerical investigation, which was totally futile.
Some other tests that I happen to know, which were inconclusive.
How might this be done?
Update: This is closer to an answer. (I think the series converges.)
Update 2: The series does in fact converge. In a comment to the original question here, Andres Caicedo mentioned this earlier identical question on MSE, which includes a rigorous answer.
The possibility that this series diverges comes from the fact that $\sin k$ regularly gets very close to one.
Given a small $\epsilon>0$, suppose $\sin k>1-\epsilon$ for at least one value of $k$ in every block of one million integers. Then $$\sum_{k=1}^{\infty}\frac{1}{k}\left( \frac{\sin{k} + 2}{3} \right) ^k> \sum_{m=1}^{\infty}\frac{1}{1000000m}\left( \frac{3-\epsilon}{3} \right) ^{1000000m}=-\frac{\log(1-(1-\epsilon)^{1000000})}{1000000}.$$
We can be precise about how often $\sin k>1-\epsilon$ (on average) by finding the function $N(\epsilon)$ for which it’s true that $\sin k>1-\epsilon$ for on average one integer in $N(\epsilon)$.
First, $\sin k = \sin(k \pmod {2\pi})$, and because $\pi$ is irrational, the real numbers $k \pmod {2\pi}$ are uniformly distributed on the interval $[0,2\pi)$. The value of $\sin x$ exceeds $1-\epsilon$ for those $x$ values between $\arcsin(1-\epsilon)$ and $\pi-\arcsin(1-\epsilon)$, which accounts for the fraction $\frac{\arccos(1-\epsilon)}{\pi}$ of the interval.
In other words, the fraction of integers $k$ for which $\sin k>1-\epsilon$ is $\frac{\arccos(1-\epsilon)}{\pi}$ and, on average, $\sin k>1-\epsilon$ for one integer in $\frac{\pi}{\arccos(1-\epsilon)}$.
I assume (but I don’t know the justification) that this asymptotic result is enough to justify the “1 in a million” argument above (at worst with $\alpha N(\epsilon)$ for some $\alpha$ slightly less than $1$). If so,
$$\sum_{k=1}^{\infty}\frac{1}{k}\left( \frac{\sin{k} + 2}{3} \right) ^k>-\frac{\log(1-(1-\epsilon)^{\frac{\pi}{\arccos(1-\epsilon)}})}{\frac{\pi}{\arccos(1-\epsilon)}}$$
for all $\epsilon>0$.
If
$$-\lim_{\epsilon\to0^+}\frac{\log(1-(1-\epsilon)^{\frac{\pi}{\arccos(1-\epsilon)}})}{\frac{\pi}{\arccos(1-\epsilon)}}=\infty,$$
the original series diverges.
However, this limit appears to be zero, or so says Mathematica. This can probably be derived using the approximation $\arccos(1-x)\approx \sqrt{2x}$ near $x=0$.
I think the failed approach to show divergence can be turned into a proof of convergence as follows, but I find it worrisome that the actual value of $N(\epsilon)$ doesn’t seem to be important.
From before, we know that for $\epsilon>0$, $\sin k\le1-\epsilon$ for the proportion $1-\frac{1}{N(\epsilon)}$ of integers. Then I think it’s the case that
$\begin{align}\sum_{k=1}^{\infty}\frac{1}{k}\left( \frac{\sin{k} + 2}{3} \right) ^k&<\sum_{m=0}^{\infty}\frac{1}{N(\epsilon)m+1}\left( \frac{3-\epsilon}{3} \right)^{N(\epsilon)m+1}+\sum_{k=1}^\infty\frac{1}{k}\left(\frac{3-\epsilon}{3} \right) ^{k}\\ &=-\frac{\log(1-(1-\epsilon)^{N(\epsilon)})}{N(\epsilon)}+\log 3-\log\epsilon. \end{align}$
A tight upper limit isn't required, so choose any nonzero $\epsilon$ and assume no more about $N(\epsilon)$ than that it’s a number (even though we know its value) to get an upper bound for the sum. I know this is a bit sloppy, but hopefully it helps and will get streamlined.
Not a solution but an idea that might help you out: $$\sum_{k \geq 1} \cfrac{1}{k}\left( \cfrac{\sin k + 2}{3} \right)^k = \sum_{k \geq 1} \cfrac{1}{3^kk}\left( \sin k + 2 \right)^k=\sum_{k \geq 1} \cfrac{1}{3^kk} \left(\sum_{l = 0}^k {k \choose l}\sin^lk\cdot2^{k-l} \right).$$ Now, $\sin k \leq 1$, so the sum is less than or equal $$\sum_{k \geq 1} \cfrac{1}{3^kk} \left(\sum_{l = 1}^k {k \choose l}2^{k-l} \right)=\sum_{k \geq 1} \frac{1}{k}\left(\cfrac{2}{3}\right)^k\left( \sum_{l = 0}^k {k \choose l} \cfrac{1}{2^l} \right)= \\ \sum_{k \geq 1} \frac{1}{k}\left(\cfrac{2}{3}\right)^k\left( \sum_{l = 0}^k \cfrac{k!}{l!(k-l)!} \cfrac{1}{2^l} \right) = \sum_{k \geq 1} \left(\cfrac{2}{3}\right)^k\left( \sum_{l = 0}^k \cfrac{(k-1)!}{l!(k-l)!} \cfrac{1}{2^l} \right).$$ If you manage to bound the inner sum (I haven't figured it out, I must confess; but at first glance it seems doable) then you are done.
