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For this series, find the radius of convergence and write it as a geometric series and give a formula if $x>3$

$\sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n$

Now finding the radius of convergence wasn't too difficult and I'll save you guys the trouble of doing it because the interval of convergence is $ -1 = 3 - 4< x < 4 + 3 = 7$ and so the radius of convergence is 2

The second part confuses me because I don't understand (if it is even possible) to convert a power series to a geometric series.

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    Yes, I made a careless mistake. Thank you for catching that Gerry again.2012-06-18

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For this particular function, shouldn't one note that $ \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n = \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n}}(x-3)^n = \frac{1}{2}\sum^{\infty}_{n=0}\left(\frac{x-3}{2}\right)^{n}$ which is a geometric series!

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    The sum then simplifies to $\frac{1}{5-x}$.2014-10-15
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It is not always possible to convert a power series to a geometric series, but in this case it can be done. A geometric series is $a+ar+ar^2+\dots$, all you have to do is a little pattern matching to figure out what $a$ and $r$ have to be in your example.