Find the limit of Riemann sums
$$\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\cos(\frac{k\pi}{2n})\frac{
\pi}{2n}$$ on the interval $$[0,\frac{\pi}{2}]$$
Progress
All I have managed to do is pull out the constant $$\frac{
\pi}{2n}\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\cos(\frac{k\pi}{2n})$$But otherwise I am completely lost. I have one question on Riemann Sums and I know the basic rules but I still cannot figure this out. The whole Riemann Sum concept still perplexes me so I thought of someone could do this problem step by step to use as guidance when dealing with these sums. Thanks for all the help in advance.
EDIT: I now realize the constant I pulled out isn't really a constant so I am even more dumbfounded. Sorry if there isn't enough work/insight on the problem form my part, but Riemann Sums are really like a different language to me. Thanks again for all the help in advance.
Best Answer
The right endpoint Riemann sum for $\displaystyle\int_a^b f(x)\,dx$ is given by $\dfrac{b-a}{n}\displaystyle\sum_{k = 1}^{n}f\left(a+\dfrac{b-a}{n}k\right)$.
Now, figure out what $a,b$ and $f(x)$ need to be to make this sum look like the one in the problem.
Once you do that, if $f(x)$ is continuous, then $\displaystyle\lim_{n \to \infty}\dfrac{b-a}{n}\sum_{k = 1}^{n}f\left(a+\dfrac{b-a}{n}k\right) = \int_a^b f(x)\,dx$.