Another way to express $\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}$? I believe that the sum $$\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}$$ converges and it is about $1.85193$.  Is there another way that this number can be expressed without summation notation. If not, perhaps there is a good approximation?
 A: $\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)
=\sin\left(\frac{2\pi n}{2m+2}\right)
=\sin\left(\frac{\pi n}{m+1}\right)
$
so
$\frac1{n}\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)
=\frac1{n}\sin\left(\frac{\pi n}{m+1}\right)
=\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)
=\frac{\pi}{m+1}\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)
$.
Therefore
$\begin{array}\\
\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}
&=\sum_{n=1}^m\frac{\pi}{m+1}\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)\\
&=\frac{\pi}{m+1}\sum_{n=1}^m\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)\\
&=\frac{\pi}{m+1}\sum_{n=1}^m\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)\\
&=\frac{\pi}{m+1}\sum_{n=1}^m\frac{\sin\left(\frac{\pi n}{m+1}\right)}{\frac{\pi n}{m+1}}\\
&\to \int_0^{\pi} \frac{\sin(t)dt}{t}
\qquad\text{as } m \to \infty\\
&=1.8519370519824...\\
\end{array}
$
according to Wolfy.
A: This is not an answer but it is too long for a comment.
Using a CAS, I found that
$$S_m=\sum_{n=1}^m \frac{e^{2 i \pi n \left(1+\frac{1}{2 m+2}\right) }}{n}=-\left(e^{\frac{i \pi }{m+1}}\right)^{m+1} \Phi \left(e^{\frac{i \pi
   }{m+1}},1,m+1\right)-\log \left(1-e^{\frac{i \pi }{m+1}}\right)$$ where appears  the Lerch transcendent function.
However, I have not been able to write the limit when $m\to \infty$. However, for very large $m$, it seems that the asymptotics is $$S_m\simeq\Phi \left(e^{\frac{i \pi }{m+1}},1,m+1\right)+\log(\frac{im}\pi) $$ For $m=10^4$, the value of the imaginary part is $\approx 1.851937049$ as you already noticed.
According to inverse symbolic calculators, this last number seems to be $$\text{Si}(\pi ) \approx 1.8519370519824661704$$ This let me suppose that there could be a rigorous proof of  $$\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}=\text{Si}(\pi )$$
Edit
Marty Cohen gave the proof.
Notice that $\frac{2 \text{Si}(\pi )}{\pi }$ is the  Gibbs constant
