Express $\frac{1}{(1-x)^2}$ as a power series by differentiating the equation below. What is the radius of convergence?

Neetesh Kumar | December 10, 2024

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This is the solution to Math 132
Assignment: 11.9 Question Number 5
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Solution:

We start with the equation:
$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \displaystyle\sum_{n=0}^\infty x^n, \quad \text{for } |x| < 1$

Differentiating each side of the equation, we get:

$\frac{1}{(1-x)^2} = 1 + \boxed{2x} + {3x^2} + \cdots = \displaystyle\sum_{n=1}^\infty \boxed{n x^{n-1}}$

Rewriting the series by replacing $n$ with $n+1$:

$\frac{1}{(1-x)^2} = \displaystyle\sum_{n=0}^\infty \boxed{(n+1)x^n}$

According to this theorem, the radius of convergence of the differentiated series is the same as the radius of convergence of the original series, namely, $R = \boxed{1}$.

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