例22将下列函数展开成x的幂级数 (1)f(x)=e2 解因f(x)=e(n=0,1,2,…)→f(0)=1(n=0,1,2,…) 于是∑x=∑1x且其收敛区间为(-∞2+) =0h! 6xn+1 n+1 而0≤R2(x) (n+1)!(m+1)! n+1 o x n+1 且 是∑ 的一般项,则对x∈( (n+1)!m=(n+1)! 恒有limR(x)=0→imR(x)=0 →0 故c2=∑ 1+x+一+一+…+—+ n 2!3! (-∞<x<+0)9 例22 将下列函数展开成x的幂级数: (1) ( ) x f x e = ( ) ( ) ( 0,1,2, ) n x 解 因 f x e n = = ( ) (0) 1 ( 0,1,2, ) n  = = f n ( ) 0 0 (0) 1 ! ! n n n n n f x x n n   = =  = 且其收敛区间为(－∞,+ ∞) 1 1 0 ( ) ( 1)! ( 1)! x n x n n e x e x R x n n  + +   =  + + 而 1 1 ( 1)! ( 1)! 0 n n n x x n n + +  + + = 且 是 的一般  恒有 项, 则对 于是   − + x ( , ), 2 3 0 1 ! 2! 3! ! n n x n x x x x e x n n  = 故 = = + + + + + +  ( ) −    + x lim ( ) 0 n n R x → =  lim ( ) 0 n n R x → =