<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] Note that $\displaystyle\left|\sum_{k=1}^n \frac{\sin k}k - \frac12(\pi-1)\right| = \left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|$.
We prove the following inequality:
$$
\left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|
\leq 
\frac{1}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
$$

Let $\displaystyle S_n = \sum_{k=1}^n \sin k = \frac{\cos(1/2)-\cos(n+\frac 12)}{2\sin(\frac 12)}$ and
$\displaystyle T_n 
= \sum_{k=1}^n S_k 
= \frac{n}{2\tan(\frac 12)} - \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)}$.
Performing summation by parts twice yields
$$\sum_{k=n+1}^\infty \frac{\sin k}k 
= \sum_{k=n+1}^\infty \Big(\frac{S_k}{k(k+1)}\Big) - \frac{S_n}{n+1}
=\sum_{k=n+1}^\infty \Big(\frac{2T_k}{k(k+1)(k+2)}\Big) - \frac{T_n}{(n+1)(n+2)} - \frac{S_n}{n+1}.
$$
Note that
$$\sum_{k=n+1}^\infty \frac{2T_k}{k(k+1)(k+2)} = \frac{1}{\tan(\frac 12)} \frac{1}{n+2} - \frac{1}{2 \sin^2(\frac 12)} \sum_{k=n+1}^\infty \frac{\sin(k+1)-\sin(1)}{k(k+1)(k+2)} $$
and
$$\begin{align}
\frac{T_n}{(n+1)(n+2)} + \frac{S_n}{n+1}
= &\phantom{-}\Big(\frac{1}{\tan(\frac 12)}-\frac{\cos(n+\frac 12)}{2 \sin(\frac 12)} \Big)
\frac 1{n+1} \\ &- 
\Big(\frac{1}{\tan(\frac 12)} +  \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)} \Big) \frac 1{(n+1)(n+2)}.
\end{align}
$$
Therefore
$$\begin{aligned}
\sum_{k=n+1}^\infty \frac{\sin k}k = 
&\phantom{+}\frac{\cos(n+\frac 12)}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
\\
&-\frac{1}{2 \sin^2(\frac 12)} \sum_{k=n+1}^\infty \frac{\sin(k+1)-\sin(1)}{k(k+1)(k+2)}
\end{aligned}
$$
and
$$\begin{aligned}
\left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|
&\leq \frac{|\cos(n+\frac 12)|}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{|\sin(n+1)-\sin (1)|+1}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)} \\
&\leq 
\frac{1}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
\end{aligned}
$$
</s>