Entonces s(x)=√
=∑ lt; -∑ lt; n=1, ∞ gt; nx^(n-1)
=[∑ lt; ; n=1, ∞ gt; x^n]'
=[x^2/(1-x)]''-[x/(1-x)]'
= 2/(1-x)^3-1/(1-x)^2 =(1 x)/(1-x)^3(-1 lt; x lt1)
Entonces σ< n = 1, ∞ gt; n^2/2^(n-1)= s(1/2)=(3/2)/(1/8)= 12
Entonces s(x)=√
=∑ lt; -∑ lt; n=1, ∞ gt; nx^(n-1)
=[∑ lt; ; n=1, ∞ gt; x^n]'
=[x^2/(1-x)]''-[x/(1-x)]'
= 2/(1-x)^3-1/(1-x)^2 =(1 x)/(1-x)^3(-1 lt; x lt1)
Entonces σ< n = 1, ∞ gt; n^2/2^(n-1)= s(1/2)=(3/2)/(1/8)= 12