Question 1089120
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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 = {{{(1-a)}}} + {{{(1-a^2)/2!}}} + {{{(1-a^3)/3!}}} + . . . = 

        = {{{1 + 1/2! + 1/3! + . . . )}}} - ({{{a}}} + {{{a^2/2!}}} + {{{a^3/3!}}} + . . .) = 

        = {{{e}}} - {{{(e^a-1)}}} = {{{1 + e - e^a}}}.


Hence,  S = {{{(1 + e - e^a)/(1-a)}}}, under the condition a =/= 1.


        At a = 1, then the sum S = 1 + e.
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Solved.