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 = + = − + − =