Homework Help: Analysis - Convergence of n ^ 1/n

1. Sep 16, 2011

pcvt

1. Prove that n^(1/n) converges to 1.

3. I've attempted to define {a} = n^1/n - 1 and have shown, using the binomial formula, that n=(1+a)^2>=1+[n(n-1)/2]*a^2. I think I'm on the right track but don't know how to bring this back to the original problem to prove convergence even after staring at it for quite a long time. Thanks!

2. Sep 16, 2011

micromass

Try to use the following trick

$$n^{1/n}=e^{\log(n)/n}$$

Now, can you figure ouy where $\frac{\log(n)}{n}$ converges to?

3. Sep 16, 2011

LCKurtz

Try working with

$$\ln( n^{\frac 1 n})$$

4. Sep 16, 2011

pcvt

Hmm our class hasn't formally introduced logs yet so I don't know if it would be kosher or not to prove it using the log.

5. Sep 16, 2011

icystrike

I'm pretty sure u are familiar with ur binomial series:

$$\left( \left( n-1\right) +1\right) ^{\frac{1}{n}}=\left( n-1\right) ^{\frac{% 1}{n}}+\frac{1}{n}\left( n-1\right) ^{\frac{1}{n}-1}+\frac{\frac{1}{n}\left( \frac{1}{n}-1\right) }{2!}\left( n-1\right) ^{\frac{1}{n}-2}+...$$

Since every term has 1/n except the first term, what conclusion can you draw?

6. Sep 16, 2011

pcvt

Wouldnt that just say n^1/n converges to (n-1)^1/n which is trivial?

I'm trying to manipulate the binomial formula, archimedean property, and bernoullis inequality to somehow get it but still haven't been able to after a couple hours

7. Sep 16, 2011

icystrike

no its not trivial. So what can u say about (n-1)^(1/n) ?

It will tend to (n-2)^(1/n) ...
It will tend to (n-3)^(1/n) ...
It will tend to (n-4)^(1/n) ...
which eventually tends to (n-(n-k))^(1/n) = k^(1/n).

Therefore ?

8. Sep 17, 2011

Ray Vickson

This argument does not work. It is true that you can conclude n^(1/n) and (n-j)^(1/n) have the same limit, for any fixed j, but you cannot continue this for j varying with n. If you _could_ ,and so have lim n^(1/n) = lim k^(1/n), then you would get 0 in the limit. But, the OP is asked to prove that the limit = 1.

RGV

9. Sep 17, 2011

icystrike

Uhh. it will tend to 1 instead of 0. since it is k^0 = 1

Secondly, n-k is some integer smaller than n therefore the argument is plausible.

10. Sep 17, 2011

Ray Vickson

Sorry: you are right. However, plausible is not the same as justifiedor provable. The argument is still invalid.

RGV

11. Sep 17, 2011

awkward

Let $f(n) = n^{1/n}$
and define $x = \lim f(n)$
See if you can show that
$$x^2 = \lim n^{2/n}$$
and
$$x = \lim f(n) = \lim f(n/2) = \lim n^{2/n}$$
so
$$x^2 = x$$
Then what can you conclude?

12. Sep 17, 2011

awkward

Somehow I thought I could justify the step above earlier, but I no longer see how to justify it. So please disregard my post. (Blush)

13. Sep 17, 2011

icystrike

Its not just plausible it has been justified if you feel uncomfortable with the former argument, u can expand (k+(n-k))^(1/n) such that k>n/2 then repeat the similar argument.

Last edited: Sep 18, 2011
14. Sep 18, 2011

icystrike

$$\lim_{x\rightarrow 0}n^{\frac{1}{n}}$$

$$=\lim_{x\rightarrow 0}\left( k+\left( n-k\right) \right) ^{\frac{1}{n}}$$

$$k>\frac{n}{2},\left\vert k\right\vert >\left\vert n-k\right\vert$$

$$=\lim_{x\rightarrow 0}\left( k^{\frac{1}{n}}+\frac{1}{n}F\left( n-k\right) \right)$$

$$=1+0$$

$$=1$$

15. Sep 18, 2011

Ray Vickson

k^(1/n) = [n + (k-n)]^(1/n) = n^(1/n)*(1 - x)^(1/n), where x = (n-k)/n. For fixed k and for n --> infinity, x --> 1-, so 1-x --> 0. Showing (1-x)^(1/n) --> 1 is not trivial. We could try a binomial expansion, so F(x) = (1-x)^(1/n) = 1 - (1/n)x + (1/n)(1/n -1)x^2 /2 + ... = 1 - G, where G = (n-k)/n^2 + [(n-1)/n^2](n-k)^2/n^2] / 2 +... . Now, it is true that each term of G is O(1/n), but G is an infinite sum, so it is not clear that we can change the limiting and summation order.

Much, much easier is the suggestion of looking at L= lim n^(1/n) and showing L = lim (2n)^(1/(2n)), to give L = lim 2^(1/(2n)) * lim sqrt(n^(1/n)) and showing 2^(1/(2n)) --> 1, so that L = sqrt(L). Then we just need some way of showing L > 0, hence L = 1.

In a way, it does not make sense to dis-allow use of the logarithm because, in fact, the very problem of showing the existence of, and properties of nth roots needs some fairly sophisticated tools. Even showing the existence of the limit, let alone finding its value, needs some work, and using logs is by far the most straightforward way.

RGV