Alt Series test

• April 7th 2009, 02:23 PM
saiyanmx89
Alt Series test
$\sum^ \infty_ {n = 1} \frac{(-1)^n}{ln(n+1)}$

How do I use the Alt series test on this summation? I'm pretty sure that the function is > 0, but I'm not sure if its decreasing.
• April 7th 2009, 02:47 PM
CaptainBlack
Quote:

Originally Posted by saiyanmx89
$\sum^ \infty_ {n = 1} \frac{(-1)^n}{ln(n+1)}$

How do I use the Alt series test on this summation? I'm pretty sure that the function is > 0, but I'm not sure if its decreasing.

Is $\ln(n+1)$ increasing?

CB
• April 7th 2009, 02:48 PM
coolguy99
Well, you can't get a negative number with ln(anything) as long as there are no negatives outside the function, so it is indeed positive. And it is decreasing, because as n increases, the denominator increases, so the values are getting smaller and smaller.
• April 7th 2009, 04:57 PM
saiyanmx89
once I took the limit of the absolute value of the function, I found it to be divergent. Am I correct? I hope so....
• April 7th 2009, 05:01 PM
mr fantastic
Quote:

Originally Posted by saiyanmx89
once I took the limit of the absolute value of the function, I found it to be divergent. Am I correct? I hope so....

Please state the alternating series test (here is some help: Alternating series test - Wikipedia, the free encyclopedia)

Now look at the previous posts again.
• April 7th 2009, 05:19 PM
saiyanmx89
1) $\frac{1}{ln(n+1)}> 0$
2) $\frac{1}{ln(n+1)}$ is decreasing
3) $\lim_{n \rightarrow \infty} \frac{1}{ln(n+1)} = \lim_{n \rightarrow \infty} (n+1)$ = $\infty$

By Alt Series Test, $\sum^ \infty _{n=1} \frac{(-1)^n}{ln(n+1)}$ is divergent.
• April 7th 2009, 05:21 PM
mr fantastic
Quote:

Originally Posted by saiyanmx89
1) $\frac{1}{ln(n+1)}> 0$
2) $\frac{1}{ln(n+1)}$ is decreasing
3) $\lim_{n \rightarrow \infty} \frac{1}{ln(n+1)} = \lim_{n \rightarrow \infty} (n+1)$ = $\infty$ Mr F says: How do you figure this?

By Alt Series Test, $\sum^ \infty _{n=1} \frac{(-1)^n}{ln(n+1)}$ is divergent.

The limit in 3) is equal to zero. The
Quote:

$= \lim_{n \rightarrow \infty} (n+1)$
bit is totally wrong.