Alternating series

**Shananay** · January 31st 2010, 10:12 PM

I'm trying to show that this series diverges.

$<br /> \sum_{n=1}^{\infty} (-1)^n\frac{1+4^n}{1+3^n}<br />$

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.

**General** · January 31st 2010, 10:21 PM

Originally Posted by Shananay

I'm trying to show that this series diverges.

$<br /> \sum_{n=1}^{\infty} (-1)^n\frac{1+4^n}{1+3^n}<br />$

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 $\color{blue}\lim_{n\to\infty} |a_n| \neq 0$ then $\color{blue}\lim_{n\to\infty} a_n$ D.N.E
Hence, $\color{blue}\sum an$ diverges by the nth term test (The test for divergence).
your $\color{blue}a_n$ is $\color{blue}(-1)^n\frac{1+4^n}{1+3^n}$ .

**Shananay** · January 31st 2010, 10:33 PM

Originally Posted by General

If $\color{blue}\lim_{n\to\infty} |a_n| \neq 0$ then $\color{blue}\lim_{n\to\infty} a_n$ D.N.E
Hence, $\color{blue}\sum an$ diverges by the nth term test (The test for divergence).
your $\color{blue}a_n$ is $\color{blue}(-1)^n\frac{1+4^n}{1+3^n}$ .

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 $\frac{\infty}{\infty}$ , but using L'Hopital, I was still getting the same answer.

**General** · January 31st 2010, 10:41 PM

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 $\frac{\infty}{\infty}$ , but using L'Hopital, I was still getting the same answer.

When you do L`Hospital`s Rule you will face:
$\color{blue}\lim_{n\to\infty} \frac{4^n ln(4)}{3^n ln(3)}=\frac{ln(4)}{ln(3)}\lim_{n\to\infty} (\frac{4}{3})^n=?$

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

**Shananay** · January 31st 2010, 10:44 PM

Originally Posted by General

When you do L`Hospital`s Rule you will face:
$\color{blue}\lim_{n\to\infty} \frac{4^n ln(4)}{3^n ln(3)}=\frac{ln(4)}{ln(3)}\lim_{n\to\infty} (\frac{4}{3})^n=?$

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

Perfect, thank you

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