Show $\sum_{l=0}^\infty \sum_{i=0}^\infty x^{{2^l}(2i+1)}=\sum_{j=1}^\infty x^j$ 
Show $\sum_{l=0}^\infty \sum_{i=0}^\infty x^{{2^l}(2i+1)}=\sum_{j=1}^\infty x^j$ where $|x|<1$

Thus show it is the geometric series.
I know, I could just write out the sum and it would make sense, but is there a more formal proof for this equality?
 A: If $|x| < 1$,
$$\sum_{i=0}^\infty x^{2^l(2i+1)} = x^{2^l} \sum_{i=0}^\infty x^{2^{l+1} i} = \frac{x^{2^l}}{1 - x^{2^{l+1}}}.$$
Now, intuitively, if you sum the first $L$ terms you get the geometric series except for terms where the power of $x$ is divisible by $2^{L+1}$.  So, you would expect the $L$th partial sum of the double sum to be equal to:
$$\sum_{l=0}^L \sum_{i=0}^\infty x^{2^l(2i+1)} = \frac{x}{1-x} - \frac{x^{2^{L+1}}}{1 - x^{2^{L+1}}}.$$
You can now either prove this by induction, or observe that it's in the form of the result of a telescoping sum, so it's sufficient to show
$$\frac{x^{2^l}}{1 - x^{2^{l+1}}} = \frac{x^{2^l}}{1 - x^{2^l}} - \frac{x^{2^{l+1}}}{1 - x^{2^{l+1}}}.$$
Then, once you have this result, it's easy to take the limit as $L \to \infty$, again assuming $|x| < 1$.
A: Rewrite your LHS as
$$
\sum_{\alpha\in A}x^{\alpha}
$$
where
$$
A:=\{2^l(2i+1)\;:\;l,i\in\Bbb N_0\}
$$
and observe that
$$
A=\Bbb N
$$
where $\Bbb N$ and $\Bbb N_0$ are natural numbers starting from $1$ and $0$ respectively.
