why $ \sum_{k=0}^{\infty} x^{2k} = \frac{1}{1-x^2}\\$ Why $$ \sum_{k=0}^{\infty} x^{2k} = \frac{1}{1-x^2}\\$$ 
I know that $$ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}\\$$
can I use the above to derive the first result? 
 A: One way you can see that this result is true with a substitution. Let $y = x^2$. Then $$ \begin{align}\sum_{k=0}^{\infty} x^{2k} =  \sum_{k=0}^{\infty} y^{k}  \\ = \frac{1}{1-y} \\ = \frac{1}{1-x^2}\end{align}$$ Making substitutions like this will become immensely useful when you get to Taylor Series.
A: Hint:
$$ \sum_{k=0}^{\infty} x^{2k} =\sum_{k=0}^{\infty} (x^2)^k $$
A: Consider the sum of even powers,
$$ \sum_{k=0}^{\infty} x^{2k}$$ and multiply it by $1+x$. You get the sum of all powers,
$$(1+x)\sum_{k=0}^{\infty} x^{2k}= \sum_{k=0}^{\infty}\left(x^{2k}+x^{2k+1}\right)=\sum_{k=0}^{\infty} x^{k}.$$
Then
$$\sum_{k=0}^{\infty} x^{2k}=\frac1{(1-x)(1+x)}.$$
A: This is simply a matter of logic. You know that $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}.$
Of course, this is just shorthand notation for the more formal:
$$\mathop{\forall}_{x:\mathbb{R}}\left(|x|<1 \rightarrow \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}\right)$$
If this notation makes no sense to you, have a look at this wikipedia page.
Anyway, by uniform substitution, we deduce:
$$\mathop{\forall}_{x:\mathbb{R}}\left(|x^2|<1 \rightarrow \sum_{k=0}^{\infty} (x^2)^k = \frac{1}{1-x^2}\right)$$
But given $x \in \mathbb{R}$, the following hold:


*

*$|x^2|<1 \iff |x|<1$

*$(x^2)^k = x^{2k}$


Therefore:
$$\mathop{\forall}_{x:\mathbb{R}}\left(|x|<1 \rightarrow \sum_{k=0}^{\infty} x^{2k} = \frac{1}{1-x^2}\right)$$
