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31 March, 03:55

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

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  1. 31 March, 04:24
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    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
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