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

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

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  • $\begingroup$ can you use $$\LaTeX$$ please? $\endgroup$ Jan 8, 2017 at 7:25
  • $\begingroup$ is this your formula here $$\sum_{m=1}^{\infty}{2}^{-2m}=1/4\sum_{m=0}^{\infty}{4}^{-m}$$ $\endgroup$ Jan 8, 2017 at 7:25
  • $\begingroup$ @Sonnhard yes thats correct.. $\endgroup$
    – Curious
    Jan 8, 2017 at 7:28
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    $\begingroup$ 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. $\endgroup$ Jan 8, 2017 at 7:31

1 Answer 1

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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 $\frac{1}{4}$: $$\sum_{m=1}^\infty (\frac{1}{4})^m = \sum_{m=0}^\infty (\frac{1}{4})^{m+1} =\frac{1}{4}\sum_{m=0}^\infty (\frac{1}{4})^m=\frac{1}{4}\sum_{m=0}^\infty 4^{-m}$$

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