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

Explanation:

Step1: Differentiate the geometric - series

The geometric - series is $\frac{1}{1 - x}=\sum_{n = 0}^{\infty}x^{n}=1 + x+x^{2}+x^{3}+\cdots$, 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 with a different index

Let $m=n - 1$, then $n=m + 1$. When $n = 1$, $m = 0$. So $\sum_{n = 1}^{\infty}nx^{n - 1}=\sum_{m = 0}^{\infty}(m + 1)x^{m}$.

Step3: Determine the radius of convergence

According to the theorem, the radius of convergence of the differentiated power series is the same as the radius of convergence of the original power series. 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 $R = 1$.