Determining convergence of infinite series with factorial without ratio test

[emr13]

Homework Statement

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.)

Homework Equations

n/a

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?

 

[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.

 

[Dick]

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

 

[emr13]

You can use the limit comparison 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.

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.

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

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

 

[Dick]

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

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

 

[emr13]

Nope.

 

[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.

 

[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.

 

[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.

 

[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.