Express 1 / (1 - x) 2 as a power series by differentiating the equation below. What is the radius of convergence? 1 1 - x = 1 + x + x2 + x3 + = [infinity] xn n = 0 |x| < 1

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