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Convergence test: Factorials

  • Thread starter Joshk80k
  • Start date
  • #1
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Homework Statement



Show that

[tex]\sum \frac{n!}{10^n}[/tex]

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:

[tex]|\frac{a_n_+_1}{a_n}|[/tex]

The Attempt at a Solution



[tex] a_n=\frac{n!}{10^n},

a_n_+_1\frac{(n+1)!}{10^n^+^1} = \frac{(n+1)n!}{10^n^+^1}[/tex]

Applying the ratio test,

[tex]|\frac{a_n_+_1}{a_n}|= \frac{10^n(n+1)n!}{n!10^n^+^1} [/tex]

Cancelling terms out,

[tex]\frac{n+1}{10} = r[/tex]

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!
 

Answers and Replies

  • #2
33,164
4,848

Homework Statement



Show that

[tex]\sum \frac{n!}{10^n}[/tex]

(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:

[tex]|\frac{a_n_+_1}{a_n}|[/tex]

The Attempt at a Solution



[tex] a_n=\frac{n!}{10^n},

a_n_+_1\frac{(n+1)!}{10^n^+^1} = \frac{(n+1)n!}{10^n^+^1}[/tex]

Applying the ratio test,

[tex]|\frac{a_n_+_1}{a_n}|= \frac{10^n(n+1)n!}{n!10^n^+^1} [/tex]

Cancelling terms out,

[tex]\frac{n+1}{10} = r[/tex]

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.
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!
 
  • #3
17
0
Alright, so I was right - it diverges.

Thanks very much!
 

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