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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!

Arctic Char
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Marc
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    It looks good. A few words about why $$\sum_{j=1}^n\frac 1{\sqrt j} < 1 + \int_1^n \frac{dx}{\sqrt x}$$ would make it easier to follow, since the reader doesn't have to stop and figure it out themselves, then remember where they were in the main proof before proceeding. But even without that it is fine. – Paul Sinclair Jul 29 '20 at 01:24
  • @PaulSinclair Point well taken. Thank you very much for checking this ! – Marc Jul 29 '20 at 01:37

1 Answers1

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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.

Simply Beautiful Art
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