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

jsewell94
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Homework Statement



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!
 
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well it is just e-1
 
I am aware that the answer is e-1, but I need to know how to get that. :)
 
jsewell94 said:
I am aware that the answer is e-1, but I need to know how to get that. :)

For the comparison part here's a hint: 1/(1*2*3)<1/(1*2*2).
 
Oh, wow...I was totally making this way more difficult than it needed to be. Thanks!
 
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?
 
jsewell94 said:
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?

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