TITLE: Convergence of recursively defined sequence
QUESTION [8 upvotes]: I have the following problem:

Suppose $a$ and $k$ are positive reals and $ a^2 > 2k $.
Set $x_{0} = a$ and define $x_{n}=x_{n-1} + \frac{k}{x_{n-1}}$ for $n\geq1$. Prove that $\lim_{n \to \infty}\frac{x_{n}}{\sqrt{n}}$ exists and determine its value.

For reference this is a problem in Hardy and William's The Green Book of Mathematical Problems.
I believe I have a proof of this and would appreciate if anyone could check for its correctness. I'd also appreciate if someone could give a simpler argument.
Here is my proof:

We show that
$$\limsup_{n \to \infty}\frac{x_{n}}{\sqrt{n}}=\liminf_{n \to \infty}\frac{x_{n}}{\sqrt{n}}=\sqrt{2k}$$
This will show the sequence converges to $\sqrt{2k}$.
From squaring the defining relation, we get
$$x_{n}^2=x_{n-1}^2+2k+\frac{k^2}{x_{n-1}^2} > x_{n-1}^2 +2k$$
Applying this same estimate $n$ times gives us the lower bound
$ x_{n}^2 > a^2+2nk$, so that $x_{n} > \sqrt{2nk+a^2}>\sqrt{2k(n+1)}$   (*)
We then get
$$\liminf_{n \to \infty}\frac{x_{n}}{\sqrt{n}} \geq \liminf_{n \to \infty}\sqrt{2k\frac{n+1}{n}}=\sqrt{2k}$$
We'll now show
$$\limsup_{n \to \infty}\frac{x_{n}}{\sqrt{n}}\leq\sqrt{2k}$$
We first derive an upper bound on $ x_{n}-x_{n-1} $.
Applying the estimate (*), to $x_{n-1}$ gives us
$$x_{n}= x_{n-1} + \frac{k}{x_{n-1}} < x_{n-1} + \frac{k}{\sqrt{2k(n-1)+a^2}} < x_{n-1} +\sqrt{\frac{k}{2n}}$$
where $a^2 > 2k$ was used.
Thus, for $n\geq1$, we get
$$x_{n}-x_{n-1} < \sqrt{\frac{k}{2n}}$$ (**)
Observe that by telescoping, we have
$$x_{n}=a+\sum_{j=1}^{n}x_{j}-x_{j-1}$$
Applying the estimate (**) to each term in the summand gives
$$\frac{x_{n}-a}{\sqrt{n}}=\frac{1}{\sqrt{n}}\sum_{j=1}^{n}x_{j}-x_{j-1}<\sqrt{\frac{k}{2n}}\sum_{j=1}^{n}\frac{1}{\sqrt{j}}$$
Estimating this last term with an integral gives:
$$\sqrt{\frac{k}{2n}}\sum_{j=1}^{n}\frac{1}{\sqrt{j}} < \sqrt{\frac{k}{2n}}(1+\int_{1}^{n}\frac{dx}{\sqrt{x}})=\sqrt{\frac{k}{2n}}(2\sqrt{n}-1)$$
Thus, we get
$$\frac{x_{n}}{\sqrt{n}}<\frac{a}{\sqrt{n}}-\sqrt{\frac{k}{2n}}+\sqrt{2k}$$
Taking the limsup of both sides gives
$$\limsup_{n \to \infty}\frac{x_{n}}{\sqrt{n}}\leq\sqrt{2k}$$
which completes the proof.
Feedback and or corrections are much appreciated!

REPLY [3 votes]: The proof looks largely to be correct. Here is an alternative approach to proving the same which uses mostly the same steps, but with some simplifications:
First note that the lower bound can be simplified from $x_n^2>a^2+2nk$ to $x_n^2>2nk$.
This can then be substituted into the squared relation to get
$$x_{n+1}^2<x_n^2+2k+\frac{k^2}{2nk}=x_n^2+2k+\frac k{2n}$$
which is much nicer to work with.
As you had noted, one may then get a bound such as
$$x_n^2<x_1^2+2nk-2k+\sum_{i=1}^n\frac k{2i}$$
by telescoping.
Dividing both sides by $n$ and taking $n\to\infty$, one gets
$$\limsup_{n\to\infty}\frac{x_n^2}n\le2k$$
where
$$\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac k{2i}=\lim_{n\to\infty}\frac k{2n}=0$$
is a Cesàro limit. This avoids the need to explicitly bound the summation as you had done.