Proving $a_n = \sum_{k=1}^{n}\frac1{k}\sin(\frac{k\pi}{n+1})$ is increasing Is the sequence $a_n = \sum_{k=1}^{n}\frac1{k}\sin(\frac{k\pi}{n+1})$ increasing?
Note that it converges to the integral $\int_0^\pi\frac{\sin x}{x} \,dx$. I was plotting this sequence on Desmos and apparently it is monotone. But I'm having an hard time proving this.
It seems that it's difficult to manipulate the sum: I don't think I can employ functions here and study some derivative; I used the sine sum-to-product formula while trying to show that $a_n < a_{n+1}$, but it makes things more cumbersome. What could be a way to start?
EDIT: for the sake of my problem, it suffices to show that the integral is un upper bound though. I'm not really interested whether it is increasing or not.
As suggested by a user, this question was asked to complete the proof given in the answer to another question, which I posted here Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$. Basically, what I'm trying to show is that the sequence of the maxima of the peaks of the partial sums of the Fourier series for the sawtooth wave is increasing to the integral given above (related to the Gibbs's phenomenon).
 A: Let $S_n(x) = \sum_{k=1}^n \frac{\sin kx}{k}$ denote the partial sum. We have
$$S_n'(x) =  \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2}\cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}, $$
and in $[0,\pi]$ the local maxima and minima of $S_n(x)$ are at
$ \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$ and $\frac{2m}{n}\pi$ for  $m = 1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$, respectively.
Note that $S_{n+1}(x) - S_n(x) = \frac{\sin (n+1)x}{n+1}$ and, thus, $S_{n+1}\left(\frac{2m+1}{n+1}\pi \right) = S_n\left(\frac{2m+1}{n+1}\pi  \right) $.
Note that $2m \leqslant n-1 < n$ and
$$\frac{2m+1}{n+2} \pi< \frac{2m+1}{n+1} \pi < \frac{2m+2}{n+2} \pi,$$
that is, the point $\frac{2m+1}{n+1} \pi$ lies between the adjacent relative maximum and minimum of $S_{n+1}$ so that
$$S_{n+1}\left(\frac{2m+1}{n+2}\pi \right) > S_{n+1}\left(\frac{2m+1}{n+1}\pi  \right) =  S_n\left(\frac{2m+1}{n+1}\pi  \right)$$.
In particular, for $m = 0$ we have
$$S_{n+1}\left(\frac{\pi}{n+2} \right) > S_{n}\left(\frac{\pi}{n+1}  \right)$$
as was to be shown.
