$\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{2n-1} f(k/n) = \int_{0}^{2} f(x)dx$

Question

$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{2n-1} f(k/n) = \int_{0}^{2} f(x)dx$. Recently, I came across this equality connecting sum and integrals. Is this true? If it is, then Is there a general equality?

Answer 1

Yes it is true. From the definition of Riemann integral, if $f$ is Riemann integrable over $[a,b]$, then
$$ \lim_{n\to \infty} \frac{b-a}{n} \sum_{k=1}^{n} f \left(a + \left(\frac{b-a}{n}\right)k \right) = \int_{a}^{b} f(x) dx$$

The LHS of the above thing says that we divided up the interval $[a,b]$ into $n$ equal intervals and took the sum of things of the form $f(a+ (\frac{b-a}{n})k) \frac{b-a}{n}$ and then let the number of intervals go to infinity.

Now, let’s say $a=0$ and $b=2$ and we divide the the interval $[0,2]$ into $2n$ intervals, then we have $$\lim_{n\to \infty} \frac{2-0}{2n} \sum_{k=1}^{2n} f \left( 0+ \frac{2-0}{2n} k\right) = \int_{0}^{2} f(x) dx \\ \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^{2n} f \left( \frac{k}{n} \right) =\int_{0}^{2} f(x) dx$$

You must observe that when the function is integrable $k$ can go from $0$ to $n-1$ or it can go from $1$ to $n$, both expressions become equivalents.

Hope it helps!