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?


$\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) $.


$\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.

  • $\begingroup$ You gave the proof I was hoping ! Thanks. $\endgroup$ – Claude Leibovici Nov 27 '15 at 6:38
  • $\begingroup$ That's a very affectionate name $\endgroup$ – qwr Nov 27 '15 at 6:47

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 $$\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}=\text{Si}(\pi )$$


Marty Cohen gave the proof.

Notice that $\frac{2 \text{Si}(\pi )}{\pi }$ is the Gibbs constant


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