What is $\lim_{N\to\infty}\frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin(n\frac{N\pi}{N+1})$? Here is what I have so far:
$$\lim_{N\to\infty} f_N \left(\frac{N\pi}{N+1}\right)$$
$$f_N (x) = \frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin(nx)$$
$$\implies \lim_{N\to\infty} f_N \left( \frac{N\pi}{N+1}\right)=\lim_{N\to\infty}\frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin\left(n\frac{N\pi}{N+1}\right)$$
$$=\lim_{N\to\infty}\frac{-2}{\pi}\left(-\sin\left(\frac{N\pi}{N+1}\right)+\frac{1}{2}\sin\left(\frac{2N\pi}{N+1}\right)-\frac{1}{3}\sin\left(\frac{3N\pi}{N+1}\right)+\cdots \pm \frac{1}{N}\sin\left(\frac{N^2 \pi}{N+1}\right) \right)$$
$$=\frac{2}{\pi}\lim_{N\to\infty}\frac{N\pi}{N+1}\left(\frac{\sin( \frac{N\pi}{N+1})}{\frac{N\pi}{N+1}} -\frac{\sin( \frac{2N\pi}{N+1})}{\frac{2N\pi}{N+1}}+\frac{\sin( \frac{3N\pi}{N+1})}{\frac{3N\pi}{N+1}} ... \pm \frac{\sin( \frac{N^2 \pi}{N+1})}{\frac{N^2 \pi}{N+1}}\right)$$
And in the last step I suppose I use the Riemann sum using midpoints to find the corresponding integral and evaluate it however I am a bit confused as to what integral I get.
If there is a better way to evaluate this limit, I am open to suggestions.
The expected answer is approximately 1.18.

Since
$$
\sin \left( {n\frac{{N\pi }}{{N + 1}}} \right) = \sin \left( {\pi n - \frac{{\pi n}}{{N + 1}}} \right) = ( - 1)^{n+1} \sin \left( {\frac{{\pi n}}{{N + 1}}} \right),
$$
we have
$$
 \frac{2}{{\pi (N + 1)}}\sum\limits_{n = 1}^N {\frac{{N + 1}}{n}\sin \left( {\frac{{\pi n}}{{N + 1}}} \right)}  \to  \frac{2}{\pi }\int_0^1 {\frac{\sin (\pi x)}{x}dx} = 1.1789797\ldots .
$$
