example video example express 1/(1 - x)^2 as a power series by differentiating the equation below. what is…

Explanation:

Step1: Differentiate the power - series

The power - series of $\frac{1}{1 - x}=\sum_{n = 0}^{\infty}x^{n}$, for $|x|\lt1$. Differentiating term - by - term using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $\frac{d}{dx}(\frac{1}{1 - x})=\frac{1}{(1 - x)^{2}}$ and $\frac{d}{dx}(\sum_{n = 0}^{\infty}x^{n})=\sum_{n = 1}^{\infty}nx^{n - 1}=1 + 2x+3x^{2}+\cdots$.

Step2: Rewrite the series

If we replace $n$ by $n + 1$ in $\sum_{n = 1}^{\infty}nx^{n - 1}$, let $m=n - 1$, then $n=m + 1$. The series becomes $\sum_{m=0}^{\infty}(m + 1)x^{m}$.

Step3: Determine the radius of convergence

The radius of convergence of a power - series and its derivative are the same. Since the radius of convergence of $\frac{1}{1 - x}=\sum_{n = 0}^{\infty}x^{n}$ is $R = 1$, the radius of convergence of $\frac{1}{(1 - x)^{2}}=\sum_{n = 0}^{\infty}(n + 1)x^{n}$ is also $R = 1$.

Answer:

The power - series of $\frac{1}{(1 - x)^{2}}$ is $\sum_{n = 0}^{\infty}(n + 1)x^{n}$ and the radius of convergence $R = 1$.