Radius of convergence of $\sum_{k=0}^{\infty} (-1)^kx^{2k}$ Problem: Find the radius of convergence of $\sum_{k=0}^{\infty} (-1)^kx^{2k}$
My initial problem with this is to re-write this as a power series. Expanding the series looks like
$-x^2 + x^4 -x^6 + x^8 + ...$. 
Any hints on how to write this is a power series i.e. of the form $\sum_{k=0}^{\infty} a_kx^k$ appreciated.
Edit: The issue is that I am trying to use the definition from my notes that the radius of convergence $R = lim_{n\rightarrow \infty} \sup \frac {1} {|a_k|^{1/k}}$
 A: Intuitively, think about what types of $x$ make this converge. $x=2$?: $$1-4+16-64+\ldots$$
Probably not. We get giant oscillations. Try for other values of $x$ as well to get a feel for it. Here are $3$ tests that could be used: 
Ratio Test: $$\lim_{k\to\infty}\left\vert\dfrac{a_{k+1}}{a_k}\right\vert=\lim_{k\to\infty}\left\vert\dfrac{(-1)^{k+1}x^{2k+2}}{(-1)^{k}x^{2k}}\right\vert=\lim_{k\to\infty}\left\vert x\right\vert^2<1\implies -1<x<1$$
Make sure to check whether $x=1$ and $x=-1$ converge or diverge. 
Root Test
$$\lim_{k\to\infty}\left\vert\sqrt[k]{x^{2k}}\right\vert=\vert x\vert^2<1$$
Geometric Series: $(-1)^kx^{2k}=(-x^2)^k\implies \vert -x^2\vert<1$ (because our $r$ is just $-x^2$ in the geometric series).
You could probably also use alternating series test, but that's not as easy.
A: If $|x|>1$, the sum does not converge because its terms don't tend to $0$.
If $|x|<1$, then the sum is absolutely convergent, because
$$\sum_{k=0}^N |x|^{2k}=\frac{1-|x|^{2N+1}}{1-|x|}<\frac{1}{1-|x|}$$
Which does not depend on $N$, thus the sum of absolute values is bounded, hence convergent.
Therefore, the radius of convergence of your series is $1$.
A: Hint: $$\sum_{k=0}^{\infty} (-1)^kx^{2k} = \sum_{k=0}^{\infty} (-x^2)^{k}$$
Do you know the radius of convergence $r$ for $\sum_{k=0}^\infty x^k$?  It converges for $-r < x < r$.  So the above series converges when $-r < -x^2 < r$.  
