Alt Series test

**saiyanmx89** · April 7th 2009, 02:23 PM

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

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.

**CaptainBlack** · April 7th 2009, 02:47 PM

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

**coolguy99** · April 7th 2009, 02:48 PM

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.

**saiyanmx89** · April 7th 2009, 04:57 PM

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

**mr fantastic** · April 7th 2009, 05:01 PM

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.

**saiyanmx89** · April 7th 2009, 05:19 PM

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.

**mr fantastic** · April 7th 2009, 05:21 PM

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

$= \lim_{n \rightarrow \infty} (n+1)$

bit is totally wrong.

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