Answer:
S₂=1/(1-x)² = (1/x)∑ n*xⁿ
the power series converges also for |r|<1
Step-by-step explanation:
we know that the infinite geometric series
S= ∑ rⁿ , from n=0 to n=∞
S= ∑ rⁿ = 1/(1-r)
then differentiating S with respect to r
dS/dr=d[1/(1-r)] = 1/(1-r)²
and
dS/dr=d(∑ rⁿ)/dr =∑ d(rⁿ)/dr = ∑ n*rⁿ⁻¹ = (1/r)∑ n*rⁿ
then
S₂= (1/r)∑ n*rⁿ = 1/(1-r)²
thus
r²*S₂- r*S₂ = ∑ n*rⁿ⁺¹-∑ n*rⁿ = ∑ (n+1-1)*rⁿ⁺¹ - ∑ n*rⁿ = (∑ (n+1)*rⁿ⁺¹ - ∑ n*rⁿ )- ∑rⁿ⁺¹ = (n+1)*rⁿ⁺¹ - (1-rⁿ⁺²)/(1-r)
r²*S₂- r*S₂ = r*S₂*(r-1) = [(n+1)*rⁿ⁺¹] - (1-rⁿ⁺²)/(1-r)
S₂ = (1-rⁿ⁺²)/[r*(1-r)²] - [(n+1)*rⁿ⁺¹ ]/[r*(1-r)]
thus |r|<1 in the first term to converge and for the second term for n→∞ , S₂= 1/(1-r)² → (n+1)*rⁿ⁺¹=n*rⁿ= 1 → ln r= -(ln n) /n
then knowing that lim (ln x)= ln (lim x)
lim (ln r)= -lim [(ln n) /n ] = - lim ( 1/n) = 0 = ln r → r=1
therefore for the second term |r|< 1
then S₂ converges for |r|< 1
