Power Series and Functions — Question 6

*Question 6 Use the differentiated geometric identity to find a power series for $x/(1-x)^2$. State the index, radius, and exact interval of convergence.

*Question 6 – Solution

For $|x|<1$, $$\frac1{(1-x)^2}=\sum_{n=1}^{\infty}n x^{n-1}.$$

$$\boxed{\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}n x^n}.$$ Multiplication by $x$ shifts every exponent upward by one but does not change the radius.

The radius remains $R=1$. At $x=1$, the terms are $n$; at $x=-1$, they are $n(-1)^n$. Neither term sequence approaches zero.

The interval of convergence is $\boxed{(-1,1)}$.