# Homework Help: Determining convergence of infinite series with factorial without ratio test

1. Jan 28, 2010

### emr13

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

Determine whether the series below is convergent or not:

$$\sum 7*\frac{n!}{n^{n-10}}$$

n=8 and the series goes to infinity
(Sorry, I couldn't get the formatting correct.)

2. Relevant equations
n/a

3. The attempt at a solution

Well, originally I thought the series was divergent, and used the nth Term Test to prove it. Except that gave me infinity over infinity. I stopped there because I can't use L'Hospital's rule because I don't know how to derive factorials.

We use a computer program to enter answers, and it told me the series didn't diverge. So I know it converges. However, I have no idea how to prove that, and an explanation is required with my answer.

I know that the Ratio test is used for series with factorials, but we have not been taught that yet. The Alternating Series doesn't apply, and I don't think the p-series test applies directly.

So that leaves the Comparison and the Limit Comparison tests. What series should I use to compare? And is there any way to get rid of the factorial?

2. Jan 28, 2010

### Bohrok

You can use the [STRIKE]limit comparison[/STRIKE] ratio test to get rid of the factorial and although I didn't do it all, it looks like it will show that the series converges.

Last edited: Jan 29, 2010
3. Jan 29, 2010

### Dick

n!<n^n, isn't it? Do you have the integral test?

4. Jan 29, 2010

### emr13

So would I use $$\frac{1}{n!}$$ as my comparison series? And if so, how do I prove that converges? Because the nth Term test doesn't prove something converges if the limit as n goes to infinity is zero.

And I don't know how to take the antiderivative of n! so I don't think the intergral test applies.

5. Jan 29, 2010

### Dick

What I was thinking of would only apply to n!/(n^(n+10)) anyway. You don't know Stirling's approximation by any chance?

6. Jan 29, 2010

### emr13

Nope.

7. Jan 29, 2010

### Bohrok

I'm sorry, I meant ratio test before.
Originally I tried to find a good comparison, but every larger series I tried didn't converge... then I started trying a limit comparison.

8. Jan 29, 2010

### Bohrok

A limit comparison with n!/n^n, which converges, has a limit of infinity. It's not zero, but does an infinite limit mean the other sum converges if one converges? I'm not having much luck finding the answer to that on Google.

9. Jan 29, 2010

### Dick

Here's one way to do it. I'm just going to do this schematically, ok? First show n!/n^n is less than roughly (1/2)^(n/2). Do this by writing out the product term by term and showing about half the terms are less then 1/2 and the rest are between 1/2 and 1. Now n!/n^(n-10) is less than ((n-10)!/n^(n-10))*(n^10). So we've got (1/2)^((n-10)/2)*n^10. At this point a ratio test would be GREAT. But let's avoid it. Use l'Hopital to show n^10 is less than than 2^((n-10)/4) for sufficiently large n. So your series is bounded by a geometric series. Awkward, you bet. It's all I could come up with.

Last edited: Jan 30, 2010
10. Feb 1, 2010

### emr13

Thanks so much for all your help.

Turns out we learned the Ratio Test just in time for me to submit the assignment, so that's what I used.