Convergence of a series (quite frustrating) 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? 
 A: 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.
A: 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.
