How can I find the interval of convergence for this power series?

Question

The problem is to find the radius of convergence and interval of convergence for the summation from $n=1$ to $\infty$ of:

$\sum ((4-x)^n)/(4^n+9^n)$

I applied the root test and got that $|x-4| < 13 = R$, so I'm almost there. I need to check the end points to see if the series converges, but this is where I am having trouble. If I test x = -9, I will have the summation of $15^n/(4^n+9^n)$, but I don't see any easy way to test for the convergence/divergence of this.

How do I proceed from here?

Answer 1

We have

$$u_n(x)=(4-x)^n \frac{1}{9^n}\frac{1}{1+(\frac{4}{9})^n},$$

and $\lim_{n\to\infty}(\frac{4}{9})^n=0$

thus

$|u_n(x)| $ is equivalent near $\infty$, to

$$\frac{|(4-x)^n|}{9^n}$$

Cauchy root test gives $\frac{|4-x|}{9}$ as a limit.

so,

if $|4-x|<9 \; \sum u_n(x) $ converges

and if $|4-x|>9 \; $, it diverges.

The convergence radius is $R=9$.

Answer 2

We have $${15^n\over 4^n+9^n} >{15^n\over 13^n}$$

You can then apply the nth root criteria :

$$ \left({15^n\over 13^n}\right)^{1\over n} = {15\over 13} > 1$$

So $\sum{15^n\over 13^n}$ does not converge and by comparaison of two positive series, $\sum {15^n\over 4^n+9^n} $ doesn't converge.