$\displaystyle \sum^ \infty_ {n = 1} \frac{(-1)^n}{ln(n+1)}$

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.

Printable View

- Apr 7th 2009, 02:23 PMsaiyanmx89Alt Series test
$\displaystyle \sum^ \infty_ {n = 1} \frac{(-1)^n}{ln(n+1)}$

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. - Apr 7th 2009, 02:47 PMCaptainBlack
- Apr 7th 2009, 02:48 PMcoolguy99
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.

- Apr 7th 2009, 04:57 PMsaiyanmx89
once I took the limit of the absolute value of the function, I found it to be divergent. Am I correct? I hope so....

- Apr 7th 2009, 05:01 PMmr fantastic
Please state the alternating series test (here is some help: Alternating series test - Wikipedia, the free encyclopedia)

Now look at the previous posts again. - Apr 7th 2009, 05:19 PMsaiyanmx89
1) $\displaystyle \frac{1}{ln(n+1)}> 0$

2) $\displaystyle \frac{1}{ln(n+1)}$ is decreasing

3) $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{ln(n+1)} = \lim_{n \rightarrow \infty} (n+1)$ = $\displaystyle \infty$

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