Ratio test works 'backwards'

Given a power series - $\sum a_n z^n$ we can use the ratio test to say this converges absolutely when lim$| (a_{n+1})/(a_n) z | < 1$ - e.g. the series has ROC lim $|a_n / a_{n+1}|$

In my notes I have 'you can use it backwards' followed by this example from the board- assume R is greater than zero and not infinity - :

given a power series (as above) has ROC R, then

$\sum (a_n)^2 z^{2n}$ will have ROC root R and use this example to find $\sum (a_n / n!) z^n$

That's what the lecturer wrote down, though I don't understand why we can use it backwards. I have tried proving we can, but haven't got anywhere because I have no clue what's going on. I imagine abs conv is used in the argument though I'm unsure.

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theres nothing backwards about it that i can see. $\lim|a_n/a_{n+1}|=R$ iff $\lim|a_n^2/a_{n+1}^2|=R^2$ –  yoyo Jan 8 '12 at 15:09
I agree, nothing backwards about it ; its just a deduction –  Patrick Da Silva Jan 8 '12 at 15:16
I missed a two and added the other bit I missed out after. I also think the root R should be a fourth root, but I am unsure if he made a mistake with the ^2n or the missed the root out (I seem to have marked in both then scribbled them out). –  Adam Jan 8 '12 at 15:17
Careful: if the limit of $|a_n/a_{n+1}|$ exists then the ROC is that limit. But there's nothing that says that $|a_n/a_{n+1}|$ must have a limit, or even a lim sup. –  Robert Israel Jan 8 '12 at 18:14