6. -/4 points details my notes scalcet9 11.9.ae.004. example video example express 1/(1 - x)^2 as a power…

Explanation:

Step1: Differentiate the power - series

We know that if $f(x)=\sum_{n = 0}^{\infty}x^{n}=1 + x+x^{2}+x^{3}+\cdots$ for $|x|<1$. Differentiating term - by - term using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $\frac{d}{dx}\left(\sum_{n = 0}^{\infty}x^{n}\right)=\sum_{n = 1}^{\infty}nx^{n - 1}$. And $\frac{d}{dx}(1 + x+x^{2}+x^{3}+\cdots)=1+2x + 3x^{2}+\cdots$. So, $\frac{1}{(1 - x)^{2}}=1 + 2x+3x^{2}+\cdots=\sum_{n = 1}^{\infty}nx^{n - 1}$.

Step2: Rewrite the series

Let $m=n - 1$, then $n=m + 1$. When $n = 1$, $m = 0$. So, $\sum_{n = 1}^{\infty}nx^{n - 1}=\sum_{m = 0}^{\infty}(m + 1)x^{m}$. We can replace $m$ with $n$ and get $\frac{1}{(1 - x)^{2}}=\sum_{n = 0}^{\infty}(n + 1)x^{n}$.

Step3: Find the radius of convergence

The radius of convergence of the power series $\sum_{n = 0}^{\infty}x^{n}$ is $R = 1$. According to the theorem that the radius of convergence of the differentiated series of a power series is the same as that of the original power series, the radius of convergence of $\sum_{n = 0}^{\infty}(n + 1)x^{n}$ is also $R = 1$.

Answer:

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