# Convergence/Divergence of a Series

1. Jul 11, 2008

### sharkshockey

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

$$\Sigma^{\infty}_{n=0}$$$$\frac{1}{2^{n}+(\frac{1}{3})^{n}}$$

2. Relevant equations

3. The attempt at a solution
I decided to use the ratio test:

$$\frac{1}{2^{n+1}+(\frac{1}{3})^{n+1}}$$ x $$\frac{2^{n}+(\frac{1}{3})^{n}}{1}$$

And I got $$lim_{n\rightarrow\infty}$$$$\frac{1}{2 + \frac{1}{3}}$$

But I'm A) pretty sure it's wrong and B) if not, what do I do after that step?

2. Jul 11, 2008

### Defennder

You can't cancel the term $$2^n + (\frac{1}{3})^n$$ just like that since $$2^{n+1} + (\frac{1}{3})^{n+1} \ \mbox{is not} \ (2^n + (\frac{1}{3})^n) \cdot (2 + \frac{1}{3})$$

Instead express $$2^n + (\frac{1}{3})^n$$ as a fraction and do the same for the (n+1) expression as well. Something will cancel out and then you should be able to apply a certain limit rule to get the answer.

3. Jul 11, 2008

### sharkshockey

I thought I could write 3n+1 as (3)(3n). So I should combine the two so that 2n + $$\frac{1}{3^{n}}$$ = $$\frac{7}{6}$$$$^{n}$$ and then find its limit?

4. Jul 11, 2008

### sharkshockey

OH! Actually, could I write the f(x) as a fraction so I would get $$\frac{1}{\frac{7}{6}^{n}}$$ and then find its convergence/divergence through a geometric series? Or should I use an integral test?

5. Jul 11, 2008

### Defennder

3^(n+1) = 3(3^n). But that isn't what you're doing here. Just express $$2^{n+1} + \frac{1}{3^{n+1}}$$ as a single fraction. Then you can apply the ratio test easily.