Power Series — Question 10

*Question 10 Find the interval of convergence and sum function of $\displaystyle\sum_{n=0}^{\infty}(x^2)^n$. Treat it as a geometric series in $x^2$ and test $x=\pm1$.

*Question 10 – Solution

The common ratio is $r=x^2$. Therefore $$|r|<1\iff |x^2|<1\iff x^2<1\iff |x|<1.$$

At both $x=1$ and $x=-1$, we have $x^2=1$, so the series becomes $\sum1$ and diverges.

For $|x|<1$, the geometric formula gives $$\boxed{\displaystyle\sum_{n=0}^{\infty}(x^2)^n=\frac1{1-x^2}}.$$

The interval of convergence is $\boxed{(-1,1)}$. As a power series in $x$, only even powers occur, but its center is still $0$ and its radius is $1$.