1. ## Alternating series

I'm trying to show that this series diverges.

$\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{1+4^n}{1+3^n}$

I'm really drawing a blank here. I know there is a quick way to show divergence but I'm not sure how. Any help appreciated.

2. Originally Posted by Shananay
I'm trying to show that this series diverges.

$\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{1+4^n}{1+3^n}$

I'm really drawing a blank here. I know there is a quick way to show divergence but I'm not sure how. Any help appreciated.
If $\displaystyle \color{blue}\lim_{n\to\infty} |a_n| \neq 0$ then $\displaystyle \color{blue}\lim_{n\to\infty} a_n$ D.N.E
Hence, $\displaystyle \color{blue}\sum an$ diverges by the nth term test (The test for divergence).
your $\displaystyle \color{blue}a_n$ is $\displaystyle \color{blue}(-1)^n\frac{1+4^n}{1+3^n}$.

3. Originally Posted by General
If $\displaystyle \color{blue}\lim_{n\to\infty} |a_n| \neq 0$ then $\displaystyle \color{blue}\lim_{n\to\infty} a_n$ D.N.E
Hence, $\displaystyle \color{blue}\sum an$ diverges by the nth term test (The test for divergence).
your $\displaystyle \color{blue}a_n$ is $\displaystyle \color{blue}(-1)^n\frac{1+4^n}{1+3^n}$.
Thank you, that's kind of what I was thinking to do but I realized I didn't even know how to evaluate that limit. I know it's not equal to zero but how to I show that? I was getting $\displaystyle \frac{\infty}{\infty}$, but using L'Hopital, I was still getting the same answer.

4. Originally Posted by Shananay
Thank you, that's kind of what I was thinking to do but I realized I didn't even know how to evaluate that limit. I know it's not equal to zero but how to I show that? I was getting $\displaystyle \frac{\infty}{\infty}$, but using L'Hopital, I was still getting the same answer.
When you do LHospitals Rule you will face:
$\displaystyle \color{blue}\lim_{n\to\infty} \frac{4^n ln(4)}{3^n ln(3)}=\frac{ln(4)}{ln(3)}\lim_{n\to\infty} (\frac{4}{3})^n=?$

Remember that:
1-$\displaystyle \color{blue} \lim_{n\to\infty} L^n = \infty$ if $\displaystyle \color{blue}|L|>1$
2-$\displaystyle \color{blue} \lim_{n\to\infty} L^n = 0$ if $\displaystyle \color{blue}|L|<1$

5. Originally Posted by General
When you do LHospitals Rule you will face:
$\displaystyle \color{blue}\lim_{n\to\infty} \frac{4^n ln(4)}{3^n ln(3)}=\frac{ln(4)}{ln(3)}\lim_{n\to\infty} (\frac{4}{3})^n=?$

Remember that:
1-$\displaystyle \color{blue} \lim_{n\to\infty} L^n = \infty$ if $\displaystyle \color{blue}|L|>1$
2-$\displaystyle \color{blue} \lim_{n\to\infty} L^n = 0$ if $\displaystyle \color{blue}|L|<1$
Perfect, thank you