Series sum $\sum_{k = 1}^\infty\frac{1}{n}\sin(\frac{k\pi}{n})$ In Apostol Calculus I textbook (section 10.4, exercise 35e), he asks to prove the series sum: 
$$ \lim_{n \rightarrow \infty}\sum_{k = 1}^n\frac{1}{n}\sin(\frac{k\pi}{n}) = \frac{2}{\pi}$$
I tried to do it by using the previous exercise: we assume that the above is the upper step function over the function $\sin(x)$. If the function is MONOTONIC on $[a, b]$, than we can use the identity:
$$ \lim_{n \rightarrow \infty}\sum_{k = 1}^n\frac{(b - a)}{n}f(a + \frac{(b-a)k}{n}) = \int_a^bf(x)dx$$
I specifically highlighted the word monotonic! The problem is that on the interval $0 - \pi$, sine will first increase, and then decrease, which renders the formula incorrect (according to the textbook previous exercise). WITHOUT the assumption about monotonicity, the same result as Apostol's is easily achieved. However, I doubt it after drawing the step function above the sine. I think the answer is incorrect. Also, I found some answers online which actually give the same answer as Apostol.
How can this be proven?
 A: The sine function is monotonic on $[0,\pi/2]$ and $[\pi/2, \pi]$.  In the summation, this is the requirement $k < N/2$ and $k > N/2$, respectively.  One way to go with this (the most directly, but perhaps not the clearest) is
\begin{align*}
\lim_{N \rightarrow \infty} &\sum_{k = 1}^N \frac{1}{N} \sin\left( \frac{k \pi}{N} \right)  \\
 &= \begin{cases}
\lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{N} \sin\left( \frac{k \pi}{N} \right)  &, \text{$N$ odd}  \\
\lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{N} \sin\left( \frac{k \pi}{N} \right)  &, \text{$N$ even}
\end{cases}  \\
&= \begin{cases}
\lim_{N \rightarrow \infty} \left( \sum_{k = 1}^{\lfloor N/2 \rfloor} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) + \sum_{k = \lceil N/2 \rceil}^{N} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) \right)  &, \text{$N$ odd}  \\
\lim_{N \rightarrow \infty} \left( \sum_{k = 1}^{N/2} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) + \sum_{k = 1 + N/2}^{N} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) \right)  &, \text{$N$ even}
\end{cases}  \\
&= \int_0^{1/2} \sin(\pi x) \,\mathrm{d}x + \int_{1/2}^{1} \sin(\pi x) \,\mathrm{d}x  \\
&= \frac{1}{\pi} + \frac{1}{\pi}  \\
&= \frac{2}{\pi}  \text{.}
\end{align*}
Here, sine is monotonic in each sum, so we can use the previous exercise, which you seem to want to do.
A: $$\lim_{n \rightarrow \infty}\sum_{k = 1}^n\frac{1}{n}\sin(\frac{k\pi}{n}) = 
 \int _0^1 \sin (\pi x) dx = \frac{2}{\pi}$$
