<s>[INST] 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?
 [/INST] $\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.
</s>