Find the sum of infinity of the series #1+ 4/(3!)+ 6/(4!)+ 8/(5!)+...# ?

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Answer 1 · 2018-06-27T15:08:22

#S=1+4/(3!)+6/(4!)+8/(5!)+...oo=2#

Let ,

#S=1+4/(3!)+6/(4!)+8/(5!)+...oo#

#S=2/(2!)+4/(3!)+6/(4!)+8/(5!)+...oo#

Here, #n^(th)term=(2n)/((n+1)!)#

So,

#S=sum_1^oo (2n)/((n+1)!)#

#S=2sum_1^oo (n)/((n+1)!)#

#S=2sum_1^oo [((n+1)-1)/((n+1)!)]#

#S=2{sum_1^oo (n+1)/((n+1)!)-sum_1^oo1/((n+1)!)}#

#S=2{sum_1^oo1/(n!)-sum_1^oo1/((n+1)!)}................to(A)#

#S=2{(1/(1!)+1/(2!)+1/(3!)+1/(4!)+...oo)#

#color(white)(................)-(1/(2!)+1/(3!)+1/(4!)+...oo)}#

#S=2{1/(1!)}#

#S=2#

Note :

#(1)color(red)(e=1+1/(1!)+1/(2!)+1/(3!)+1/(4!)+...oo#

#(2)color(red)(e- 1=1/(1!)+1/(2!)+1/(3!)+1/(4!)+...oo=sum_1^oo1/(n!)#

#(3)color(red)(e-2=1/(2!)+1/(3!)+1/(4!)+...oo=sum_1^oo1/((n+1)!)#
From #(A)#

#S=2[(e-1)-(e-2)]=2[e-1-e+2]=2(1)=2#