# Radius of convergence of power series

1. May 5, 2013

### Alcubierre

1. The problem statement, all variables and given/known data

The coefficients of the power series $\sum_{n=0}^{∞}$$a_{n}(x-2)^{n}$ satisfy $a_{0} = 5$ and $a_{n} = (\frac{2n+1}{3n-1})a_{n-1}$ for all $n ≥ 1$. The radius of convergence of the series is:
(a) 0
(b) $\frac{2}{3}$
(c) $\frac{3}{2}$
(d) 2
(e) infinite

2. Relevant equations

3. The attempt at a solution

$lim_{n\rightarrow ∞}|\frac{a_{n+1}(x-2)^{n}}{a_{n}(x-2)^{n}}|$

$|x-2|lim_{n\rightarrow ∞} |\frac{a_{n+1}}{a_{n}}|$

To find the function inside the limit, I used the definition presented in the problem:

$a_{n} = (\frac{2n+1}{3n-1})a_{n}$

Added n + 1 and divided the $a_{n}$ from the RHS which yields,

$\frac{a_{n+1}}{a_{n}} = \frac{2n+3}{3n+2}$

and substituted that into the limit, and took the limit:

$\frac{2}{3}|x-2| < 1$

$2|x-2| < 3$

$2[-(x-2)] < 1\; and\; 2(x-2) < 2$

And that gives the interval,

$x > \frac{1}{2}\; and\; x < \frac{7}{2}$

which means the radius is 2, but the correct answer is letter c, $\frac{3}{2}$.

What am I doing wrong? Was my approach in the right direction?
I solved this before and I used the rationale that the radius of convergence is the reciprocal of the limit of the interval, but my teacher said that wasn't correct and it was "one of those special cases that it works out that way."

Last edited: May 5, 2013
2. May 5, 2013

### LCKurtz

They satisfy what???!

3. May 5, 2013

### Alcubierre

My apologies.

It should say, it satisfies $a_{0} = 5$ and $a_{n} = \frac{2n+1}{3n-1}$ for all $n ≥ 1$.

Last edited: May 5, 2013
4. May 5, 2013

### LCKurtz

That isn't what you used in your solution and it won't give the correct answer. I think you still have something typed wrong.

5. May 5, 2013

### Alcubierre

The coefficients of the power series $\sum_{n=0}^{∞}$$a_{n}(x-2)^{n}$ satisfy $a_{0} = 5$ and $a_{n} = (\frac{2n+1}{3n-1})a_{n-1}$ for all $n ≥ 1$

Sigh, it's been a long day.

6. May 5, 2013

### LCKurtz

OK. Then your work is OK at $\frac 2 3 |x-2|<1$ or $|x-2|<\frac 3 2$. That says the distance from $x$ to $2$ is less than $\frac 3 2$. Since $x=2$ is the center of the expansion $r = \frac 3 2$.

7. May 5, 2013

### Alcubierre

Oh wow I completely disregarded the fact that it's centered at x = 2! Thank you very much