# Alternating series test problem

1. Mar 14, 2014

### jdawg

1. The problem statement, all variables and given/known data

n=1∑(-1)n$\stackrel{10n}{(n+1)!}$

2. Relevant equations

3. The attempt at a solution

I already found that the limit does equal zero by using the ratio test on bn. What I'm having trouble with is determining if it decreases or not. I know you can't take the derivative of a factorial... Does an exponential increase faster than a factorial?

2. Mar 14, 2014

### jdawg

Oops! I don't know what I did wrong trying to type that the first time. It's supposed to look like this:
(-1)n((10n)/(n+1)!)

3. Mar 14, 2014

### jbunniii

FYI, to see how to typeset, right click this expression, and choose "Show Math As TeX Commands":
$$\sum_{n=1}^{\infty} \frac{(-1)^n 10^n}{(n+1)!}$$
Then bracket the commands in double-dollar signs like so:
Code (Text):
$$\sum_{n=1}^{\infty} \frac{(-1)^n 10^n}{(n+1)!}$$
Note that if the ratio test succeeds, it implies absolute convergence.

4. Mar 14, 2014

### jdawg

Ohh! Thanks so much! I've been having a lot of trouble trying to figure out how to format things correctly. So you wouldn't have to prove that it decreases? I thought that for the alternating series test you first had to prove that the limit equals zero, and then you had to prove that the series decreases?

5. Mar 14, 2014

### jbunniii

You only have to use the alternating series test if the series doesn't converge absolutely. Absolute convergence implies convergence, so there is no need for further testing. The $(-1)^n$ factor is irrelevant.

6. Mar 14, 2014

### Dick

If you have to use the alternating series test you can. If $a_n$ is the nth term of your series then you want to show $|a_{n+1}|<|a_n|$. That's the same as showing $\frac{|a_{n+1}|}{|a_n|}<1$. That's not true for every n. But it only has to be true for large n. At what value of n does it become true? If you don't have to, then as jbunniii said, the ratio test is a better choice.

Last edited: Mar 14, 2014
7. Mar 14, 2014

### jdawg

Thanks so much for the help, I think I get it now :)