# Double summation without changing the order of sigmas

I want to evaluate the following double sum (from blackpenredpen's video) $$\sum_{m=1}^{\infty}{\sum_{n=1}^{m}\frac{1}{2^{m+n}}}$$ In the video, he is showing how to change the order of the sums, but I want to do this normally.

I am pretty confident the inner sum is $$\frac{2^m-1}{2^{2m}}$$ by geometric series although I am not that good at this subject. Then, how would I evaluate the outer sum?

$$\sum_{n=1}^{m}\frac{1}{2^{m+n}}=\frac{1}{2^{m}}\sum_{n=1}^{m}\frac{1}{2^{n}}=\frac{2^m-1}{2^{2m}}$$
$$\sum_{m=1}^{\infty} \frac{2^m-1}{2^{2m}}=\sum_{m=1}^{\infty} \frac{1}{2^{m}}-\sum_{m=1}^{\infty} \frac{1}{2^{2m}}=\sum_{m=1}^{\infty} \frac{1}{2^{m}}-\sum_{m=1}^{\infty} \frac{1}{4^{m}}$$