Question is to evaluate $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=0}^{n}f(\frac{k}{n})$$
I see that this is to be done with Riemann sums..
I first partition $[0,1]$ as $\{0,\frac{1}{n},\frac{2}{n},\dots,1\}$
Now, Riemann Sum is defined as $\sum_{i=1}^n f(x^*_i)|(x_i-x_{i-1})|$
So, here i would have
$$f(0).\frac{1}{n}+f(\frac{1}{n})\frac{1}{n}+f(\frac{2}{n})\frac{1}{n}+\dots+f(\frac{n-1}{n})\frac{1}{n}=\frac{1}{n}.\sum_{k=0}^{n-1}f(\frac{k}{n})$$
Or
$$f(\frac{1}{n})\frac{1}{n}+f(\frac{2}{n})\frac{1}{n}+\dots+f(1)\frac{1}{n}\frac{1}{n}.\sum_{k=1}^{n}f(\frac{k}{n})$$
I would get first case if i take $x^*_i=x_{i-1}$ and i would get second case if i take $x^*_i=x_i$
But I am afraid I would not get $$\frac{1}{n}\sum_{k=0}^{n}f(\frac{k}{n})$$
I am actually confused..
Should the question be something like $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})$$?
I am sure i am missing something..
Please help me to clear that.
Thank you