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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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    You are confusing the radius of curvature with the radius of convergence, and you got the wrong answer.2012-06-18
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    What if we set $\dfrac{x-3} 2 $ as $r$, and take $1/2$ out?2012-06-18
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    Oh shoot, wrong term. I have no idea why i did that Gerry.2012-06-18
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    You still have the wrong answer for the radius of convergence.2012-06-18
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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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    You haven't left OP much to do, have you?2012-06-18
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    As for the radius of convergence, we need $$ \biggl| \dfrac{x - 3}{2} \biggr| < 1. $$2012-06-18
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    @Eugene, I basically also got what Alex has, but why are you assuming $\biggl| \dfrac{x - 3}{2} \biggr| < 1$ instead of $|x - 3| < 2$? I realize you can't use $1/(1 - x)$ if I don't assume that the ratio is less than 1. But when x > 3, doesn't the sum diverge?2012-06-18
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    @jak What you wrote is the same thing as what I wrote just so you know.2012-06-18
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    Oh It's because you substituted $r = \biggl| \dfrac{x - 3}{2} \biggr| < 1$. Okay gotcha2012-06-18
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    @jak The thing is, here we don't care if $a$ is complex (it is $2$) but if it was complex $|z|$ is not necessarily $= z$. You would've gotten a different result if it was $\frac{x-2}{-2}$, maybe.2012-06-18
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    @jak: the leading factor doesn't contribute to the radius of convergence, only to the value it converges to. When you write $\left| \frac {x-3}2 \right | \lt 1$ this is the same as $|x-3| \lt 2$ so the radius of convergence is 2, centered at $x=3$2012-06-18
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    @jak: but the leading $\frac 12$ is not part of the ratio. The ratio between successive terms is just $\left| \frac {x-3}2 \right|$ which is what needs to be $\lt 1$. The $\frac 12$ divides out in the ratio.2012-06-18
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    @everyoen sorry, you are right R = 2. I got 4 because I forget to carry a (1/2) when I DIVIDE.2012-06-18
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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.