tn 解2、∑ +n n!(2 lim-n+=0 收敛域(-∞,+∞) 令S(t)=∑ n! n=In! n t n(n-1)!n2(n-2)! e+ te +t 故S(x)=e21+ o+O 例、利用计算幂级数的和函数,求下列级数的和 (-) 解5(2里=m()+(-)=8+ 记:S1(x)=∑n(n-1x=x2∑n(n-1)x2 2x 2)27解 2、   +  =     +   =  n 0 n=0 n 2 n 2 t n! 1 n 2 x t 2 x n! 1 n 0 u u lim n n 1 n = + → 收敛域（-∞，+∞） 令 ( )  =  +  + =  =  =  = n 1 n 2 n 0 n n 0 n 2 t n! n n! t t n! 1 n S t ( ) ( )  − − +  = + − = +  =  = n 1 t n n 1 t n t n 1 ! n 1 1 t e n 1 ! n e ( ) ( )  −  + − = +  =  = n 2 n n 1 n t t n 2 ! 1 n 1 ! t e t t 2 t = e + te + t e 故 ( )         = + + 4 x 2 x S x e 1 2 2 x ，(−,+) 例、利用计算幂级数的和函数，求下列级数的和 ( ) n 2 n 0 n 2 n n 1 1 − +  −  = 解： ( ) ( ) 1 2 n 0 n 0 n n n 0 n 2 n S S 2 1 2 1 n n 1 2 n n 1 S 1    = +       + −       = − − − + = −  =  =  = 3 2 2 1 1 1 S2 = + = 记： ( ) =  ( − ) =  ( − )  = −  = n 2 2 n 2 n 0 n S1 x n n 1 x x n n 1 x （-1，1） ( )          − =        =  − =   =  = − 1 x x x n n 1 x x x x 2 2 x 2 2 n n 2 2 n 2 ( ) 27 4 2 1 S 1 x 2x 3 1 2  =      − − = ∴ ( ) 27 22 3 2 27 4 2 n n 1 1 n 0 n 2 n  = + = − + −  =