radius of convergence problem

May 2009
71
0
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.
 
Oct 2008
1,116
431
Maple is right. The radius of convergence is 0.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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 \(\displaystyle x\) for which

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

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

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

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

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

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

\(\displaystyle \left|\frac{x + 3}{4}\right| < 0\).


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.