HELP! Stuck on an Infinite Series. Thanks

  • Thread starter danerape
  • Start date
  • #1
32
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I have the infinite series...

[itex]\sum n^n/n![/itex]

somehow, I need to use the Ratio Test... and shoe that the resulting limit
is equal to e. I can't figure out what I am doing wrong algebraically.

Right now I have simplified this limit to lim((n+1)^n/n^n).

Please Help...


Thanks.
 
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Answers and Replies

  • #2
32
0
THE SERIES IS n^n/n!
 
  • #3
34,686
6,393
I have the infinite series...

[itex]\sum n^n/n![/itex]

somehow, I need to use the Ratio Test... and shoe that the resulting limit
is equal to e. I can't figure out what I am doing wrong algebraically.

Right now I have simplified this limit to lim((n+1)^n/n^n).
This is the same as (1 + 1/n)n.

To evaluate a limit like this, let y = (1 + 1/n)n. Now take the ln of both sides, and then take the limit as n -> infinity. Your textbook should have an example of this type of limit.
 
  • #4
32
0
I know how to evaluate this type of limit, my problem is algebraically getting to the point of writing lim((n+1)^n/n^n)=lim(1+1/n)^n. How do I go from my ratio test to seeing the limit as equal to the latter limit?


Thanks
 
  • #5
34,686
6,393
It's fairly basic algebra.
[tex]\frac{(n + 1)^n}{n^n} = \left(\frac{n+1}{n}\right)^n[/tex]
 
  • #6
32
0
Wow, third shift is really getting to me. Thanks Alot.
 
  • #7
186
1
You should all ready know this property of exponents:

[itex]\frac{a^{n}}{b^{n}}[/itex]=([itex]\frac{a}{b}[/itex])[itex]^{n}[/itex]

Bleh...too late. I fail at tex. lol
 

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