<s>[INST] Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ converges as $n$ increases, to a limit of $\frac12(\pi-1) \approx 1.0708$
Empirically, it seems to be bounded by about $\frac12(\pi-1)  \pm \frac{1.043}{n}$, as shown in the chart below.
What is the precise value of this $1.043$ term?
Added: I think the three answers from Oliver Diaz, Gary, and Gabriel Romon, plus the comments, have demonstrated that the bounds are $$\frac12(\pi-1)  \pm \frac1{2\sin(1/2)}\frac{1}{n}$$ not only asymptotically but also as tight actual bounds; $\frac1{2\sin(1/2)}\approx 1.0429148214667441$

created in R with
n <- 1:11000
plot(n, cumsum(sin(n)/n), ylim=c(1.07,1.0715), xlim=c(1,10000), pch=46)
abline(h=(pi-1)/2, col="red")
curve((pi-1)/2 + 1.043/x, from=1, to=max(n), add=TRUE, col="red")
curve((pi-1)/2 - 1.043/x, from=1, to=max(n), add=TRUE, col="red")

 [/INST] A slightly worse estimate may be obtained as follows. If $|\arg z|<\pi$ then
$$
\log (1 + z) = \sum\limits_{k = 1}^N {( - 1)^{k + 1} \frac{{z^k }}{k}}  + ( - 1)^N z^{N + 1} \int_0^1 {\frac{{t^N }}{{1 + zt}}dt} 
$$
for any $N\geq 0$. Substituting $z =  - e^i$, taking the imaginary part of each side and rearranging, we find
$$
\sum\limits_{k = 1}^N {\frac{{\sin (k)}}{k}}  - \frac{{\pi  - 1}}{2} = \Im \left[ { - e^{i(N + 1)} \int_0^1 {\frac{{t^N }}{{1 - e^i t}}dt} } \right].
$$
Here
\begin{align*}
\left| {\Im \left[ { - e^{i(N + 1)} \int_0^1 {\frac{{t^N }}{{1 - e^i t}}dt} } \right]} \right| & \le \int_0^1 {\frac{{t^N }}{{\left| {1 - e^i t} \right|}}dt}  \le \frac{1}{{\sin 1}}\int_0^1 {t^N dt} \\ & = \frac{1}{{(N + 1)\sin 1}} < \frac{{1.1884}}{{N + 1}}.
\end{align*}
This is larger than the expected result by a factor of $\frac{1}{{\cos (1/2)}}=1.13949\ldots$. An exact formula for the remainder can be obtained by computing the imaginary part explicitly, but it results in a complicated formula which does not seem to be useful for better estimates.
Addendum. From the link you posted, the remainder term is
$$
{\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)\sin \left( {\left( {n + \tfrac{1}{2}} \right)t} \right)dt}  - \int_{n + 1/2}^\infty  {\frac{{\sin t}}{t}dt} }.
$$
Integration by parts gives
$$
\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)e^{i\left( {n + \frac{1}{2}} \right)t} dt}  = ie^{i\left( {n + \frac{1}{2}} \right)} \left( {1 - \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}} \right)\frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Taking imaginary parts
$$
\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)\sin \left( {\left( {n + \tfrac{1}{2}} \right)t} \right)dt}  = \left( {1 - \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}} \right)\frac{{\cos \left( {n + \frac{1}{2}} \right)}}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
By another integration by parts
$$
\int_{n + 1/2}^\infty  {\frac{{\sin t}}{t}dt}  =  \frac{1}{n}\cos \left( {n + \tfrac{1}{2}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Thus, your remainder term is
$$
-\frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}\frac{1}{n}\cos \left( {n + \tfrac{1}{2}} \right) +  \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right),
$$
which for large $n$ is at most
$$
 \approx \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}\frac{1}{n}
$$
in absolute value. Because of the big-$\mathcal{O}$, in principle, it could be slightly larger, so this is not a complete proof.
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