# Convergence test: Factorials

1. Apr 4, 2010

### Joshk80k

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

Show that

$$\sum \frac{n!}{10^n}$$

converges or diverges.

(Note, I was unsure of how to format this via latex, so the summation is from n = 1 to infinity.)

2. Relevant equations

The root test:

$$|\frac{a_n_+_1}{a_n}|$$

3. The attempt at a solution

$$a_n=\frac{n!}{10^n}, a_n_+_1\frac{(n+1)!}{10^n^+^1} = \frac{(n+1)n!}{10^n^+^1}$$

Applying the ratio test,

$$|\frac{a_n_+_1}{a_n}|= \frac{10^n(n+1)n!}{n!10^n^+^1}$$

Cancelling terms out,

$$\frac{n+1}{10} = r$$

Now, I know that if:

r > 1, it is divergent,
r < 1, convergent,
r = 1, inconclusive.

My problem is that I am not sure where to go after this. I still have an "n" in my answer, and I expected to just have a numerical answer.

I was going to just go ahead and say that since n approaches infinity, r is greater than 1, and thus the series is divergent, but I stopped because I realized that from n = 1 to n = 8, the series would be convergent, and worse still, at n = 9, the test would be inconclusive?

So my question is, did I make a mistake somewhere here, or is the ratio test not applicable here for this reason?

Thanks for any feedback!

2. Apr 4, 2010

### Staff: Mentor

It doesn't matter what happens between 1 and 8; you're interested in the behavior for large n. Every finite series converges, since you're just adding a finite number of terms.

3. Apr 4, 2010

### Joshk80k

Alright, so I was right - it diverges.

Thanks very much!