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I was reviewing the proof the remainder estimate for a Taylor series expansion and I came across something I can't find an intuitive explanation for: if you have a function f that's bounded on an interval $[a-s, a+s]$ and define $f_1(x) := \int_a^xf(t)\,\mathrm dt$ and define $f_n(x) := \int_a^x f_{n-1}(t)\,\mathrm dx$, then $$\lim_{n\to\infty}f_n(x) = 0.$$

Can anyone explain why or how this is the case on an intuitive level?

Also if I try this iteration using $f(x) = \cos(x)$, at each iteration I get either $\sin(x)$ or $\cos(x)$ and a part of its Taylor expansion with the next iteration resulting in a better estimate.

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    Just deal with the case $f$ constant: you have $f_n(x)=\frac{x^n}{n!}$ which converges pointwise to $1$.2012-10-07
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    @Davide: You mean to $0$?2012-10-07
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    @joriki: yes, it was a typo.2012-10-07

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