Proving limit of the nth root of n

In summary, the conversation discusses proving the limit of n^(1/n) = 1 as n approaches infinity. The solution involves choosing a value for N in terms of epsilon and using the binomial theorem to expand the inequality. This results in a proof for the limit using epsilon-delta.
  • #1
Elzair
11
0

Homework Statement



Prove the following limit:

[tex]lim_{n \rightarrow \infty } n^{ 1 / n } = 1 [/tex]

Homework Equations



Not sure.

The Attempt at a Solution



Given any [tex]\epsilon > 0[/tex], choose [tex]N \in \mdseries N[/tex] s.t.

[tex]\left| n^{ 1 / n } - 1 \right| < \epsilon[/tex] for all [tex]n > N[/tex]

I am not sure how to proceed.
 
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  • #2
If you're still working on this problem and need to do it with epsilons and deltas, I think that choosing N = exp(log(1+ε)-1) should suffice. I can't find a nice/elegant epsilon delta solution to this problem, but maybe someone else can.
 
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  • #3
Prove [tex]\lim _{n\to \infty} \sqrt[n]{n}=1[/tex].
Proof: We want:
[tex]
|\sqrt[n]{n}-1|<\epsilon
[/tex]
The abs sign can be safely dropped, it follows that
[tex]
n<(1+\epsilon)^n
[/tex]
Using binomial theorem to expand the first 3 terms of RHS.
[tex]
n<1+n\epsilon+\frac{1}{2}n(n-1)\epsilon^2+...
[/tex]
As long as we make n<0.5n(n-1)εε, the first inequality holds. It requires
[tex]
n>1+\frac{2}{\epsilon^2}
[/tex]

With all that said,
For any ε>0, there exists N=[1+2/(εε)], such that if n>N, then
[tex]
|\sqrt[n]{n}-1|<\epsilon
[/tex]

Q.E.D
P.S. I love ε-δ proof:)
 
Last edited:
  • #4
This thread is 2 years old. Please be more careful before posting in an old thread.
 

1. What is the definition of a limit of a sequence?

The limit of a sequence is the value that the terms of the sequence approach as the index (n) approaches infinity. This means that as n gets larger and larger, the terms of the sequence get closer and closer to the limit value.

2. How do you prove the limit of the nth root of n?

To prove the limit of the nth root of n, we can use the squeeze theorem. This involves finding two other sequences that are both greater than or equal to, and less than or equal to, the nth root of n, and whose limits are equal. This shows that the limit of the nth root of n must also be equal to that same value.

3. What are the steps for using the squeeze theorem to prove a limit?

The steps for using the squeeze theorem to prove a limit of the nth root of n are as follows:

  • Find two other sequences that are both greater than or equal to, and less than or equal to, the nth root of n.
  • Show that the limits of these sequences are equal.
  • Apply the squeeze theorem to show that the limit of the nth root of n must also be equal to that same value.

4. Can you provide an example of using the squeeze theorem to prove a limit of the nth root of n?

Yes, let's say we want to prove that the limit of the nth root of n is equal to 1. We can use the sequences 1 and 1.1 as our upper and lower bounds, respectively. Both of these sequences are greater than or equal to, and less than or equal to, the nth root of n. Additionally, both of these sequences have a limit of 1. Therefore, by the squeeze theorem, the limit of the nth root of n must also be equal to 1.

5. Are there any other methods for proving the limit of the nth root of n?

Yes, there are other methods for proving the limit of the nth root of n, such as using the ratio test or the direct proof method. However, the squeeze theorem is often the most efficient and straightforward method for proving limits involving nth roots.

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