函數項級數

(1)ByM.I.
f(n+1)(x)=∫_[0~x] f(n)(t) dt
=1/(n-1)! ∫_[0~x] ∫_[0~t] fo(s)(t-s)^(n-1) ds dt
=1/(n-1)! ∫_[0~x]∫_[s~x] fo(s)(t-s)^(n-1) dt ds (Change theorder of integration)
=1/(n-1) ∫_[0~x] fo(s) (1/n) (x-s)^n ds
=1/n! ∫_[0~x] fo(t) (x-t)^n dt (change dummy variable s to t)

(2)
∑[1~∞] |f(n)(x)|
<= ∑[1~∞] ∫_[0~x] M(x-t)^(n-1) /(n-1)! dt
(since |fo(t)| is conti. on[0,x], |fo(t)| <= M for some M)
=∑[1~∞] Mx^n/n! =M(e^x -1)
so that ∑[n=1~∞] f(n)(x) is convergent absolutely.