Power Series and Functions — Question 3

*Question 3 Differentiate the geometric series term by term to find a power series for $1/(1-x)^2$. Reindex it in powers $x^n$ and state its full interval of convergence.

*Question 3 – Solution

For $|x|<1$, $$\frac1{1-x}=\sum_{n=0}^{\infty}x^n.$$ A power series may be differentiated term by term at every point strictly inside its radius.

$$\frac{d}{dx}(1-x)^{-1}=(1-x)^{-2}=\sum_{n=1}^{\infty}n x^{n-1}.$$ Let $k=n-1$: $$\boxed{\frac1{(1-x)^2}=\sum_{k=0}^{\infty}(k+1)x^k}.$$

Differentiation preserves the radius $R=1$, but endpoints must be retested. At $x=1$ the terms are $k+1$; at $x=-1$ their magnitudes are $k+1$. Neither tends to zero.

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