why $ \sum_{k=0}^{\infty} x^{2k} = \frac{1}{1-x^2}\\$

Question

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?

Answer 1

Hint:
$$ \sum_{k=0}^{\infty} x^{2k} =\sum_{k=0}^{\infty} (x^2)^k $$

Answer 2

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.

Answer 3

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 $$\sum_{k=0}^{\infty} x^{2k}=\frac1{(1-x)(1+x)}.$$

Answer 4

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:

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