Question

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

Answer 1

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