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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.