# Thread: Quick Power Series Question

1. ## Quick Power Series Question

the series is..

[(3^n)(x+4)^n]/(n^0.5)

did the ratio, ended up with

3(x+4)(n^.5)/(n+1)^.5

I got the interval of convergence correctly, but for the radius, the back of the book is saying 1/3 (when, what I'm having right now is 3x+12<1 ).

So my question is just.. how did they get the 1/3? Do numbers not count if they're not attached to variables or did I just make a mistake

2. Originally Posted by coolguy99
the series is..

[(3^n)(x+4)^n]/(n^0.5)

did the ratio, ended up with

3(x+4)(n^.5)/(n+1)^.5

I got the interval of convergence correctly, but for the radius, the back of the book is saying 1/3 (when, what I'm having right now is 3x+12<1 ).

So my question is just.. how did they get the 1/3? Do numbers not count if they're not attached to variables or did I just make a mistake
You did a great job. Think of radius as the maximum distance away from x. To help see this consider the following calculations.
First of all the ratio test gives us,

$|3x+12|<1 \implies -1 < 3x + 12 < 1$

$\implies -13 < 3x < -11 \implies \frac{-13}{3} < x < \frac{-11}{3}$

$\implies -(4+\frac{1}{3}) < x < -(4-\frac{1}{3}) \implies$

So you can see x can move 1/3 units in either direction from -4.

An alternate way of seeing this is from the ratio test you have

$|3(x+4)| < 1 \implies |x+4| < \frac{1}{3}$ (this is the way you'd do it on a test {quicker}).

If you cannot see it by this form, then let's go one more step
$\implies -\frac{1}{3} < x+4 < \frac{1}{3} \implies -4-\frac{1}{3} < x < -4+\frac{1}{3}$