# how to convert /understand /reason the summation conversion?

How to interpret / reason / understand the summation conversion for the below equation ?

$$\sum_{m = 1}^\infty 2^{-2m} = \frac{1}{4}\sum_{m = 0}^\infty 4^{-m}$$

• can you use $$\LaTeX$$ please? Jan 8, 2017 at 7:25
• is this your formula here $$\sum_{m=1}^{\infty}{2}^{-2m}=1/4\sum_{m=0}^{\infty}{4}^{-m}$$ Jan 8, 2017 at 7:25
• @Sonnhard yes thats correct.. Jan 8, 2017 at 7:28
• It sometimes helps to write out the first two or three terms in full to see what is going on in these kinds of situations. Jan 8, 2017 at 7:31

There are two steps here. First, pulling down the $-2$ in the exponent: $$\sum_{m=1}^\infty 2^{-2m} = \sum_{m=1}^\infty (\frac{1}{4})^m$$
Next, shifting the lower index to zero (by replacing $m$ everywhere with $m+1$) and pulling out a factor of $\frac{1}{4}$: $$\sum_{m=1}^\infty (\frac{1}{4})^m = \sum_{m=0}^\infty (\frac{1}{4})^{m+1} =\frac{1}{4}\sum_{m=0}^\infty (\frac{1}{4})^m=\frac{1}{4}\sum_{m=0}^\infty 4^{-m}$$