Suppose that $x > 1$. Prove that $s_n \to 1$

Question: Suppose $x > 1$. Prove that $x^\frac{1}{n} \to 1$.
The following is a list of what I am trying to do through out my proof.

1. Show $s_n$ is monotone (decreasing).
2. Show $s_n$ is bounded.
3. Using the monotone convergence thm and thm 19, show that $s_n$ converges to 1.
(Thm 19: If a sequence ($s_n$) converges to a real number s, then every subsequence of ($s_n$) also converges to s.)

Proof
(1) We prove $s_n$ is decreasing by induction.
Since $s_1 = x > s_2 = \sqrt{x}$
Now assume $s_k \ge s_{k+1}$, then
$$s_{k+2} = x^\frac{1}{k+2} < x^\frac{1}{k+1} = s_{k+1}$$
Therefore, $s_n$ is monotone (it is decreasing).
(2a) We prove $s_n$ is bounded above by showing $x^2$ is an upper bound.
Since $s_1 = x < x^2$ and $s_n$ is decreasing, we can conclude $s_n$ is bounded above.
(2b) We prove $s_n$ is bounded below by showing $1$ is a lower bound.
Since $s_1 = x$ and $x > 1$, we can conclude $x^\frac{1}{n} > 1$ and that $s_n$ is bounded below.
Therefore, $s_n$ is bounded.
(3) By the monotone convergence thm, we know $s_n \to s$.
Now for each n, $$s_{2n} = x^\frac{1}{2n} = {x^\frac{1}{n}}^\frac{1}{2} = \sqrt{s_n}$$.
By thm 19, lim $s_{2n} =$ lim $s_n =$ lim $\sqrt{s_n} =$ $\sqrt{lim s_n}$
Thus, $s = \sqrt{s}$
$s^2 - s = 0$
$s = 0, 1$
Since $s_1 = x > 1$ and $s_n$ is bounded below by $1$, we can conclude $s \ne 0$.
Hence, $s = 1$ and $s_n \to 1$.

Concerns: Is part (2) of my proof sufficient to show that $s_n$ is bounded?

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Kind of an off topic question, but, is there a way to make formulas appear larger? Like in part 3 of my proof, the fractions are very tiny. :( –  enlgmatlc Apr 7 '11 at 1:05
Part (2) is sufficient to show $s_n$ is bounded (depending on what facts you've already proven about $x^{1/n}$), but it's not sufficient to rule out $s = 0$ at the end. –  Jason DeVito Apr 7 '11 at 1:08
In what way? The title is the question to the problem. –  enlgmatlc Apr 7 '11 at 1:11
I don't see how you can conclude that the limit must be 1. You know its decreasing and that the limit is either 0 or 1. It starts above 1 so why can't it decrease all the way to 0? However, it's not too hard to show that 1 is a lower bound as well. –  Eric O. Korman Apr 7 '11 at 1:11
Ah, I see, I'll fix it! Thanks Jason. –  enlgmatlc Apr 7 '11 at 1:15

There is no real problem in the proof. The sequence $(s_n)$ is decreasing. To show it has a limit there is no need to show that the sequence is bounded above, so that part of the argument is superfluous.

You observed that $0$ is a lower bound. That's fine, but it is roughly as easy to show that $1$ is a lower bound. And this is actually needed later.

In the next to last line, you write that since $s_1>1$ and the sequence is decreasing, $s \ne 0$. That is not a correct argument, it is easy to come up with a decreasing sequence $(a_n)$ such that $a_1>1$ and the sequence has limit $0$. You need to say that $s_n>1$, and therefore $s \ne 0$.

By the way, induction is not needed (and in fact not used in your proof) to show that $(s_n)$ is decreasing.

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I always like to use the definition. Besides, you already have a fine critique of your current solution. Here is another.

Let $\epsilon > 0$ be arbitrarily small. Then for all $n > \log x / \log (\epsilon+1)$ (note that $\log x > 0$ since $x>1$) we have $$x < (\epsilon+1)^n,$$ which implies $$|x^{1/n} - 1| < \epsilon,$$ as required.

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$$x^{1/n} = \exp\left(\frac{\log(x)}{n}\right)$$

As $n$ increases and $\log(x)$ stays constant this tends toward $\exp(0) = 1$.

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As an alternative suggestion, you could consider the sequence $\log(s_n)$. Since $\log$ is continuous and one-to-one, $s_n\to 1$ if and only if $\log(s_n)\to 0$. But $\log(s_n) = \log(x^{1/n}) = \frac{1}{n}\log(x)$. Since $\log(x)$ is constant (and positive), you just need to examine the sequence $\frac{1}{n}$, which is easy to do.

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