Only one option is correct ,Question based on Infinite Arithmetico-Geometric Series ?

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Answer 1 · 2017-05-08T05:31:37

Answer is (a) as #s~~0.9207#

Here we are given #s=sum_(n=1)^(oo) n e^(-n)#

= #e^(-1)+2e^(-2)+3e^(-3)+4e^(-4)+5e^(-5)+...................#

Hence #es=1+2e^(-1)+3e^(-2)+4e^(-3)+5e^(-4)+6e^(-5)+...................#

Hence #se-s=1+e^(-1)+e^(-2)+e^(-3)+e^(-4)+e^(-5)+.......................#

or #s(e-1)=1/(1-e^(-1))=e/(e-1)#

and #s=e/(e-1)^2~~2.7183/2.9525=0.9207#

Hence #s<=1# and answer is (a)

Answer 2 · 2017-05-08T11:06:20

# e/(e-1)^2=0.920674#

#sum_(n=1)^oo nx^n=x sum_(n=0)^oonx^(n-1)=x d/(dx)(sum_(n=0)^oo x^n)#

Here #x = e^-1 < 1# so

#sum_(n=0)^oo x^n=1/(1-x)# then

#sum_(n=1)^oo nx^n=x/(1-x)^2# for #abs x < 1# or

#sum_(n=1)^oo n(e^-1)^n=e^-1/(1-e^-1)^2 = e/(e-1)^2=0.920674#