Infinite sequence

  • Thread starter aceetobee
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  • #1
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I want to find the smallest value of N (n > N) such that the following is true:

abs((2n)^(1/n) - 1) < 0.01

So basically, I'm looking for the term of the sequence (2n)^(1/n) which is less than 0.01 from the limit of the sequence... by trial and error, I found it to be 734.

But I'm looking ofr a theoretical solution. Any help as to what I'm missing here?

Thanks.
 

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  • #2
HallsofIvy
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So you want 0.99< (2n)n< 1.01. That's basically looking for
(2n)n= 1.01 Since that involves n both in the base and the exponent, you won't be able to find any solution in terms of elementary functions. You might be able to use "Lambert's W function":
http://en.wikipedia.org/wiki/Lambert's_W_function
 
  • #3
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aceetobee, are you trying to prove that [itex]\lim_{n \rightarrow \infty} (2n)^{\frac{1}{n}} = 1[/itex]? If so, then trying to find the smallest N for each epsilon is not necessary. If you actually do need to find the smallest N then disregard this post.

When proving limits of sequences, we don't care about the smallest possible value of N for a particular epsilon, all we care is that an N exists having the property that n>N implies (insert the rest of the limit definition here). Note if we do have an N that works, we could just as well replace it with any larger number.

To prove this limit, I would use the fact that
[tex] (2n)^{\frac{1}{n}} = (2^{\frac{1}{n}})(n^{\frac{1}{n}})[/tex]
and use the product theorem for limits.
 
Last edited:

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