sum of series n!/1000

[petrol.veem]

I'm new to the whole sequence and series game, and I'm having a terrible time solving these problems!

My latest:

Sum(0,infinity) [ n!/1000^n ]

I started out by doing the divergence test, ie see if the limit of [ a(n) ] was equal to zero or not.

So, I considered the first and last terms multiplied by one another and compared them to the overall series

(n/1000)(1/1000) < n!/1000

Then, lim(n-->infinity) n/1000^2 = infinity

And since the lesser of the two items stated above diverges, I would guess that this series diverges as well.

Is this making sense?

 

[galactus]

I think you're on the right track. Here's a way to do it.

When you have a factorial, the ratio test is a good test a lot of the time.

\(\displaystyle \lim_{n\to{\infty}}\frac{(n+1)!}{1000^{n+1}}\cdot\frac{1000^{n}}{n!}=\frac{n+1}{1000}\)

\(\displaystyle \lim_{n\to{\infty}}\frac{n+1}{1000}\)

Now, does this limit diverge.

 

[petrol.veem]

Now, does this limit diverge.

I think so... :wink: