# Showing That The Infinite Series 1/n! is less than 2

1. Apr 21, 2012

### jsewell94

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

Consider the series:
$\sum\frac{1}{n!}$, where n begins at one and grows infinitely larger (Sorry, I'm still a bit new to the equation editor on here :) )
1) Use the ratio test to prove that this series is convergent.

2) Use the comparison test to show that S < 2

3) Write down the exact value of S.

2. The attempt at a solution

The first part of this problem was rather simple.

However, parts 2 and 3 have me completely stumped. I have tried comparing $\frac{1}{n!}$ to $\frac{1}{n^2}$, but when n = 4, $\frac{1}{n!}$ becomes smaller than $\frac{1}{n^2}$. Which leads me to believe that this would be true for any series of the form $\frac{1}{n^p}$.

I have also considered using a geometric series, but, again, I can't think of any that would remain less than $\frac{1}{n!}$...

So, what exactly do I compare it too? You don't have to outright give me the answer, but a nudge in the right direction would be nice. And I figure that once I get part 2, part 3 SHOULD fall into place.

Thanks guys!

2. Apr 21, 2012

### andrien

well it is just e-1

3. Apr 21, 2012

### jsewell94

I am aware that the answer is e-1, but I need to know how to get that. :)

4. Apr 21, 2012

### Dick

For the comparison part here's a hint: 1/(1*2*3)<1/(1*2*2).

5. Apr 21, 2012

### jsewell94

Oh, wow...I was totally making this way more difficult than it needed to be. Thanks!

6. Apr 21, 2012

### jsewell94

So I managed to prove that the sum is less than 2. Now how do I go about finding the exact value of the sum?

7. Apr 21, 2012

### Dick

For that one I think you need to use the power series expansion of e^x.