Find $\lim_{n \rightarrow \infty} \frac{1}{n} \sum^n_{k=1} \sin(\frac{2k-1}{2n})$ I was thinking that you either take the limit as $n$ goes to infinity first and that ends up being zero, so the whole thing would be zero, but that doesn't account for the $2n$ in the denominator of the sum. 
On the other hand, I was trying to think of bounds for the functions. since sine is bound in $[-1,1]$ then the sum couldn't be more than $\pm n$ and that multiplied with $1/n$ gives $\pm 1$ as the limit, but limits have to be unique...

 A: The function inside the limit is the $n$-point Riemann sum of $\sin x$ where the points lie at the middle of intervals $[0,1/n],[1/n,2/n],\dots,[(n-1)/n,1]$. Thus the limit evaluates to
$$\int_0^1\sin x\,dx=[-\cos x]_0^1=1-\cos1$$
A: You are facing the sum of sines in arithmetic progression (see here). So
$$\sum^n_{k=1} \sin\left(\frac{2k-1}{2n}\right)=\frac{\sin ^2\left(\frac{1}{2}\right)} {\sin \left(\frac{1}{2 n}\right)}=2\sin ^2\left(\frac{1}{2}\right)\frac {\frac 1{2}}{{\sin \left(\frac{1}{2 n}\right)} }$$ So
$$S_n=\frac 1n\sum^n_{k=1} \sin\left(\frac{2k-1}{2n}\right)=2\sin ^2\left(\frac{1}{2}\right)\frac {\frac 1{2n}}{{\sin \left(\frac{1}{2 n}\right)} } \to 2\sin ^2\left(\frac{1}{2}\right)=1-\cos(1)$$
If you want an approximation, using Taylor series, we should get
$$S_n=\left(1-\cos(1)\right)\left(1+\frac{1}{24 n^2}+\frac{7}{5760 n^4}+O\left(\frac{1}{n^6}\right)\right)$$
Use your pocket calculator for $n=6$. The "exact" value is $0.4602301830$ while the above truncated expansion would give                       $0.4602301827$
A: Below is my naive attempt. I believe the answer will be $\int_{0}^{1}\sin x dx$. 
According to the Eric's comments, I can write $\sin(\frac{2k - 1}{2n})$ into $\sin(\frac{k}{n})\cos(\frac{1}{2n}) - \cos(\frac{k}{n})\sin(\frac{1}{2n})$. The latter one will approach to $0$ as $n$ approach $\infty$. The first term will be like $\cos(\frac{1}{2n}) \frac{1}{n}\sum_{k \leq n}\sin(\frac{k}{n})$. Hence this part will converge to $\int_{0}^{1} \sin x dx$
