Convergence/Divergence of a Series

[sharkshockey]

Homework Statement

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

Homework Equations

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?

 

[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.

 

[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?

 

[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?

 

[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.