Direct comparison test for convergence

  • Thread starter RadiationX
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  • #1
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I'm supposed to compare the series

[tex]\sum_{n=0}^{\infty}\frac{1}{n!}[/tex]

to some other series to see if the one above converges or diverges. I have no idea of what to compare it to.

I know by the ratio test that the above series converges, that is if i'm doing the ratio test correctly.

[tex]\lim_{n\rightarrow\infty}\frac{1}{(n+1)!}n!\\=\frac{1}{n(n+1)}(1*n)=\\\frac{1}{(n+1)}[/tex]

since this limit is zero which is less than one the series converges
 

Answers and Replies

  • #2
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I'm thinking about majoring in mathematics. How important are sequences and series to a math major? I hate them BTW.
 
  • #3
StatusX
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Compare it to the sum of 1/2^n. If you can show this converges, and show that every term in this series is greater than the corresponding term in the original series, that one must also converge.
 
  • #4
HallsofIvy
Science Advisor
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Or even, and simpler, just compare it to [tex]\frac{1}{n^2}[/tex]. As soon as n> 2,
[tex]\frac{1}{n^n}< \frac{1}{n^2}[/tex]. Now, what do you know about the convergence of [tex]\Sigma_1^\infty\frac{1}{n^2}[/tex]?
 

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