Let S = 1+(1+a)/2!+(1+a+a^2)/3!+....infinity Multiply all and every terms/term by (1-a). You will get (1-a)*S = + + + . . . = = - ( + + + . . .) = = - = . Hence, S = , under the condition a =/= 1. At a = 1, then the sum S = 1 + e.