Is this proof of lim((2n)^(1/n)) = 1 correct?

[Hodgey8806]

Homework Statement

I need to show the lim((2n)^(1/n)) = 1

Homework Equations

I will be using the definition of the limit as well as using the Binomial Theorem as an aide.

I am following an example from my book quite similar. So applying the Binomial Theorem to this problem, I will choose to write (2n)^(1/n) as 1 + Kn for some Kn > 0.

Raising both sides to the n power, we have 2n = (1 + Kn)^n ≥ 1 + (1/2)n(n-1)(Kn)^2
=> 2n ≥ (1/2)n(n-1)(Kn)^2
I will then solve for Kn to get that Kn≤ 2/√n-1

This tells us that there is some Nε such that 2/√Nε-1 < ε since ε>0 (By the Archimedean Property).

The Attempt at a Solution

Now, applying that to my proof, I have:
Let ε>0 be given.
0<(2n)^(1/n) -1 = (1 + Kn) -1 = Kn ≤ 2/√n-1 < ε
Since ε>0 is arbitrary, we can conclude that lim((2n)^(1/n)) = 1

I appreciate the help! I would like to know if this is mostly correct, and would like help in rewriting it to make it neater. Thank you :)

 

[mathwonk]

I don't understand your justification for beginning the way you do, by setting a complicated function like (2n)^(1/n) equal to a simple linear function like 1+Kn. After raising them both to the nth power, you can see even more clearly how unequal they are since then you have on the left the very slowly growing linear function 2n, and on the right an extremely rapidly growing function comparable to n^n, which is growing faster even than the exponential function e^n.

Your use of quantifiers (i.e. lack of them) is also quite unclear to me. Namely you choose several different quantities apparently independently but also in a combined way. I.e. I have trouble seeing when e>0 is chosen or given, and if it is given one should only choose N rather than Ne, and perhaps K rather than Kn.

In line 5, assertions such as (essentially) 2n ≥ (Kn)^2 should give pause, since the RHS is obviously much larger in general than the LHS. (Just try n = 3, 4, 5, ..., and K = 1.)

I would use l'Hopital's rule on a limit like this. Oddly enough, I do get the limit as 1. You should of course say where n is tending to before you can make sense of taking a limit. I.e. does n tend to infinity? I would start over, replace n by x, and look up l'Hopital's rule. Good luck.

 

[malawi_glenn]

What limit is this n->0 or 1 or what?

 

[malawi_glenn]

I don't understand your justification for beginning the way you do, by setting a complicated function like (2n)^(1/n) equal to a simple linear function like 1+Kn.

I think it is due to poster not using LaTeX

$(2n)^{1/n} = 1 + k_n$ where $k_n$ is some real number $k_1 = k_2 = 1$, $k_3 = 6^{1/3}- 1$

 

[mathwonk]

Yes I think I understand what is going on now. K is chosen as a function of n as you suggest. Then for each n, there is a unique positive number K(n) such that (2n)^(1/n) = 1 + nK(n).
Then 2n = (1+nK(n))^n ≥ 1 + (nK(n))^2.(n)(n-1)/2, hence 2n-1 ≥ (nK(n))^2.(n)(n-1)/2.

Dividing by n(n-1)/2, we get (4n-2)/[(n)(n-1)] ≥ (nK(n))^2.

Since the LHS goes to zero as n --> infinity, so does the right side. Hence as n goes to infinity, so does nK(n). Hence 1+nK(n) = (2n)^(1/n) ---> 1.

(And if I had fully understood your remark I would have just taken K(n) so that (2n)^(1/n) = 1+K(n), and shown K(n)--->0. Having misunderstood the OP's notation, and trying to make sense of it as it was, I was mentally picturing a linear graph y = 1 + Kx, passing through (0,1) and intersecting the graph of y = (2x)^(1/x) above the point x = n.)