radius of convergence problem

**orendacl** · May 16th 2010, 12:33 AM

n!(x+3)^n
-----------
4^n

I do all the work, and come up with some ridiculous fraction:

(n+1)(1/16)^n*|x+3|
___________________
4

Take limit as n-> inf and I get zero?

I used maple as well to check and it gets:

(n+1)!(x+3)^(n+1)*4^n
_____________________
4^(n+1)*n!(x+3)

= 1/4|(x+3)(n+1)|
take limit and it becomes infinite.

Which is right?
note:
I did not show all my steps. i used the ratio test on the original problem and as you guys know, replace all n with n+1, times by reciprocal of fraction etc, You then get what i have above. I used a calculator for mine, and then tried to duplicate it with maple but get two different answers. What I need to do is find the radius of convergence.

**alexmahone** · May 16th 2010, 12:45 AM

Maple is right. The radius of convergence is 0.

**Prove It** · May 16th 2010, 01:00 AM

Originally Posted by orendacl

n!(x+3)^n
-----------
4^n

I do all the work, and come up with some ridiculous fraction:

(n+1)(1/16)^n*|x+3|
___________________
4

Take limit as n-> inf and I get zero?

I used maple as well to check and it gets:

(n+1)!(x+3)^(n+1)*4^n
_____________________
4^(n+1)*n!(x+3)

= 1/4|(x+3)(n+1)|
take limit and it becomes infinite.

Which is right?
note:
I did not show all my steps. i used the ratio test on the original problem and as you guys know, replace all n with n+1, times by reciprocal of fraction etc, You then get what i have above. I used a calculator for mine, and then tried to duplicate it with maple but get two different answers. What I need to do is find the radius of convergence.

You need to find the values of $x$ for which

$\lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1$

$\lim_{n \to \infty}\left|\frac{\frac{(n + 1)!(x + 3)^{n + 1}}{4^{n + 1}}}{\frac{n!(x + 3)^n}{4^n}}\right| < 1$

$\lim_{n \to \infty}\left|\frac{4^n(n + 1)!(x + 3)^{n + 1}}{4^{n + 1}n!(x + 3)^n}\right| < 1$

$\lim_{n \to \infty}\left|\frac{(n + 1)(x + 3)}{4}\right| < 1$

$\left|\frac{x + 3}{4}\right|\lim_{n \to \infty}(n + 1) < 1$ .

$\left|\frac{x + 3}{4}\right| < \frac{1}{\lim_{n \to \infty}(n + 1)}$

$\left|\frac{x + 3}{4}\right| < 0$ .

It is impossible to have a size less than 0, so there is not anywhere where the series converges. In other words, the radius of convergence is 0.

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