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 &lt;- 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=&quot;red&quot;)
curve((pi-1)/2 + 1.043/x, from=1, to=max(n), add=TRUE, col=&quot;red&quot;)
curve((pi-1)/2 - 1.043/x, from=1, to=max(n), add=TRUE, col=&quot;red&quot;)
This is a twice the bound of the OP:
Suppose $c_n\searrow0$.
Fix $M\in\mathbb{N}$ and for $n\geq N$ let
$$P_n(x)=\sum^n_{k=M}e^{ikx}=\frac{e^{iMx}-e^{i(n+1)}}{1-e^{ix}}=e^{iMx}\frac{1-e^{(n-M+1)ix}}{1-e^{ix}}=\frac{\sin\big((n-M+1)x/2\big)}{\sin(x/2)}e^{i(n-M)x/2}$$
From this, we get that
$$ S_n(x)=\sum^n_{k=M}\sin kx =\frac{\sin\big((n-M+1)x/2\big)}{\sin(x/2)}\sin\big((n-M)x/2\big)
$$
Then by summation by parts
$$\begin{align}
\sum^N_{n=M} c_n e^{inx} &amp;=c_NP_N(x) +\sum^{N-1}_{n=M}(c_n-c_{n+1})P_n(x)\\
\sum^N_{n=M} c_n \sin nx &amp; = c_NS_N(x) +\sum^{N-1}_{n=M}(c_n-c_{n+1})S_n(x)
\end{align}$$
Then
$$
\begin{align}
\Big|\sum^N_{n=M} c_n e^{inx}\Big| &amp;\leq c_N|P_N(x)| +\sum^{N-1}_{n=M}(c_n-c_{n+1})|P_n(x)|\\
\Big|\sum^N_{n=M} c_n \sin nx\Big| &amp;\leq c_N|S_N(x)| +\sum^{N-1}_{n=M}(c_n-c_{n+1})|S_n(x)|
\end{align}
$$
For $P_n$ and $S_n$ we have the estimates
$$\begin{align}
|P_n(x)|&amp;= \big|\sin\big((n-M+1)x/2\big)\big|\csc(x/2)|\leq  |\csc(x/2)| \\
|S_n(x)|&amp;=\Big|\sin\big((n-M+1)x/2\big)\sin\big((n-M)x/2\big)\Big||\csc(x/2)|\leq|\csc(x/2)|
\end{align}$$
for all $x\neq0$ and $n$,
\begin{align}
\Big|\sum^N_{n=M} c_n e^{inx}\Big|&amp;\leq \Big(c_N +\sum^{N-1}_{n=M}(c_n-c_{n+1})\Big)|\csc(x/2)|\leq c_M |\csc(x/2)|\tag{1}\label{one}\\
\Big|\sum^N_{n=M} c_n \sin nx\Big|&amp;\leq \Big(c_N +\sum^{N-1}_{n=M}(c_n-c_{n+1})\Big)|\csc(x/2)|\leq  c_M|\csc(x/2)|\tag{2}\label{two}
\end{align}
For the OP's series
$$
\sum^\infty_{n=1}\frac{\sin nx}{n}=\frac{1}{2i}\sum_{|n|\geq1}\operatorname{sign}(n)\frac{e^{inx}}{n}=\frac12(\pi-x),\qquad 0&lt;x&lt;2\pi
$$
From  \eqref{one} or \eqref{two} we obtain bound
$$\Big|\sum^{N-1}_{n=1}\frac{\sin nx}{n}-\frac{\pi-x}{2}\Big|=\Big|\sum^\infty_{n=N}\frac{\sin nx}{n}\Big|\leq \frac{1}{N}|\csc(x/2)|
$$
For $x=1$, we get twice the OP's bound. I ignore at the moment how to improve the bound in \eqref{one} by a factor of $1/2$ which is what the OP is observing.
Comment: The value of this sum is indeed (pi-1)/2. This is confirmed by noting that the function f(x) = pi-x on (0, pi), extended oddly to (-pi, 0), and then extended to all real numbers periodically and with the usual convention, has Fourier series = sum(sin(nx)/n; n=1,2,...). The extended function is continuous except at integer multiples of pi. Hence, we have equality at x=1, so the sum in question = f(1) = (pi-1)/2.
For more details, see:
Michael W. Ecker, Combining Mathematics and Technology to Sum Certain Series, Mathematics and Computer Education, Winter 2013, Vol. 47, No. 1, p. 19-23.