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.
- Show $s_n$ is monotone (decreasing).
- Show $s_n$ is bounded.
- 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?
Best Answer
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.