why $\int_0^1f= \lim_{n\to+\infty}\sum_{k=2}^n\int_{\frac{1}{k}}^{\frac{1}{k-1}}f$ I'm looking for an explanation for why this equality is true assuming f is integrable. obviously it is true for a finite amount of integrals covering exactly [a,b], but how do we know it is also true for the infinite case?
My intuition is that the part of $[0,1]$ we're ignoring for each n gets smaller, meaning it converges to 0, and therefore both sides are identical. Am i right thinking this way?
If anyone cares to write a proof to convince me it will be greatly appreciated.
$$\int_0^1f= \lim_{n\to+\infty}\sum_{k=2}^n\int_{\frac{1}{k}}^{\frac{1}{k-1}}f$$

$$\sum_{k=2}^{\infty}\left(\frac{1}{k-1}\:-\:\frac{1}{k}\right) = 1$$
This sum is simply equal to $1$.
