Conquering the Conundrum: Simplifying the Sum of n!/(1000^n)

[FrostScYthe]

inf
Sum n!/(1000^n)
i = 1

Is that with geometric series... or what?!... all I can think of is splitting n! and the fraction (1/1000)^n, anyway, any help appreciated :)

 

[interested_learner]

The first question you need to ask is whether this sum converges or diverges.

Have you studied Big-O notation? Is 1000^n O(n!)? Or visa versa? Don't be fooled by what happens at small values of n. Eventually one of the expressions will overwhelm the other. Which is it? and what happens when it does?

 

[StatusX]

Do the ratio test.

 

[FrostScYthe]

n! would overwhelm 1000^n eventually, I think, if I'm correct... I am guessing it diverges.. but I need to prove this algebraically...
ratio test hmmm I'm not sure how to do this for that?

n!/n! 1
------------ = -------------- agh I don't get anywhere X\
(1000^n)/n! (1000^n)/n!

 

[d_leet]

n! would be less than 1000^{n} for all n. Since a_{n} \rightarrow 0 as n\rightarrow \infty, the series converges.

Try the ratio test on this series, I think you're mistaken.

 

[d_leet]

\lim_{n\rightarrow \infty} |\frac{(n+1)n!}{1000^{n+1}}\frac{1000^{n}}{n!}| = \frac{n+1}{1000}

As n\rightarrow \infty, |\frac{a_{n+1}}{a_{n}}| \rightarrow 0. So it converges.

How does (n+1)/1000 approach 0 as n goes to infinity?

 

[quasar987]

I'm trying to reply by quoting your post #5 courtri but the quote juste won't appear.

Anyway, I just want to point out in relation to that post, that given a series \sum a_n, it is a nessesary condition for convergence that a_n\rightarrow 0 but it is not sufficient! In other words, that some series has a vanishing general term does not means it converges.

However, if a series has a diverging general term, then it diverges. It is a useful criterion for proving a series diverges, but cannot be used as a proof for convergence.

 

[quasar987]

And also in relation to your post #5 (why did you delete it, it was an intructive error that the OP could have beneficiated from). It can be seen in a nice way from the way you wrote the sequence that it diverges:

\frac{n!}{1000^n}=\frac{1}{1000}\frac{2}{1000}...\frac{n}{1000}

As soon as n\geq 1000*1000=1000000 for instance, you can rearange the terms two by two like so (ok it looks ridiculous i admit):

\frac{1000000!}{1000^{1000000}}=\left(\frac{1}{1000}\frac{1000000}{1000}\right)\left(\frac{2}{1000}\frac{999999}{1000}\right)...\left(\frac{500000}{1000}\frac{500001}{1000}\right)

Everyone's greater than one so it is easy to see that past n=1000000, the general term is always greater than unity, so the limit will not converge to 0, so the series must diverge.

 

[matt grime]

Anyway, I just want to point out in relation to that post, that given a series \sum a_n, it is a nessesary condition for convergence that a_n\rightarrow 0 but it is not sufficient! In other words, that some series has a vanishing general term does not means it converges.

You don't mean vanish. Terms vanish if they are zero, thus you have described a finite sum. Tending to zero is not the same as vanish (being equal to zero).

 

[quasar987]

Cool, I didn't know.