Factorial Series Convergence: Investigating the Sum of n!/10^n

Joshk80k
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Homework Statement



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.)

Homework Equations



The root test:

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

The Attempt at a Solution



a_n=\frac{n!}{10^n},<br /> <br /> 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!
 
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Joshk80k said:

Homework Statement



Show that

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

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

converges or diverges.



Homework Equations



The root test:

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

The Attempt at a Solution



a_n=\frac{n!}{10^n},<br /> <br /> 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?
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.
Joshk80k said:
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!
 
Alright, so I was right - it diverges.

Thanks very much!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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