高等微積分問題

a.
Let f(x)=Σ(1~∞) [n x^n]/(n+1)!, then
f(x)=Σ(1~∞) (n+1)x^n/(n+1)! -Σ(1~∞) x^n/(n+1)!
=Σ(1~∞) x^n/n! - (1/x)Σ(1~∞) x^(n+1)/(n+1)!
=e^x -1 -[ e^x -1 -x]/x
= (1- 1/x)e^x+ 1/x
b.
Let f(x)=Σ(1~∞) (-1)^n x^(2n+1)/[(2n+1)(2n+2)!], then f(0)=0 and
f'(x)=Σ(1~∞) (-1)^n x^(2n)/(2n+2)!
=[Σ(1~∞) (-1)^n x^(2n+2)/(2n+2)! ]/x^2
=[Σ(2~∞) (-1)^n x^(2n)/(2n)!]/x^2
=[ cosx -1 + 0.5x^2]/x^2
f(x)=∫[0~x] (cost -1+0.5t^2)/t^2 dt (integration by parts)
=-(cosx -1+ 0.5x^2)/x +∫[0~x] (t-sint)/t dt
=(1-cosx)/x + x/2 -∫[0~x] sint/t dt

2010-06-15 01:52:46 補充：
謝謝!
f'(x)=Σ(1~∞) (-1)^n x^(2n)/(2n+2)!
=[Σ(1~∞) (-1)^n x^(2n+2)/(2n+2)! ]/x^2
=-[Σ(2~∞) (-1)^n x^(2n)/(2n)!]/x^2 =[ -cosx +1 - 0.5x^2]/x^2
f(x)=∫[0~x] (-cost +1-0.5t^2)/t^2 dt (integration by parts)
=(cosx -1+ 0.5x^2)/x -∫[0~x] (t-sint)/t dt
=-(1-cosx)/x - x/2 +∫[0~x] sint/t dt

2010-06-15 17:36:53 補充：
真是眼殘!
f'(x)=Σ(0~∞) (-1)^n x^(2n)/(2n+2)! =[Σ(0~∞) (-1)^n x^(2n+2)/(2n+2)! ]/x^2
=-[Σ(1~∞) (-1)^n x^(2n)/(2n)!]/x^2 =[1-cosx]/x^2
f(x)=∫[0~x] (1-cost)/t^2 dt (integration by parts)
=(-1+cosx)/x +∫[0~x] sint /t dt