Iterated Radical Expansions and Convergence

As $k\rightarrow\infty$, the recurrence

$x_{1}=1,$ $x_{k}=\sqrt{1+x_{k-1}}$ for $k\geq 2$

approaches the Golden mean [3]:

$$\begin{array}[c]{ccccc}\varphi=\dfrac{1+\sqrt{5}}{2},&&\varphi^{2}=\varphi+1,&&\dfrac{1}{\varphi}=\varphi-1\end{array}$$

from below and enjoys exponential convergence [4]:

$$0<\lim_{n\rightarrow\infty}\,(2\varphi)^{n}(\varphi-x_{n})=2\,{\displaystyle\prod\limits_{k=2}^{\infty}}\,\frac{2\varphi}{\varphi+x_{k}}<\infty.$$

We shall prove this formula using entirely elementary techniques.

First, notice that $1\leq x_{k}<\varphi$ for all $k$ by induction ($x_{k}\geq 1$ is obvious; supposing $1\leq x_{k-1}<\varphi$, we obtain

$$x_{k}=\sqrt{1+x_{k-1}}<\sqrt{1+\varphi}=\sqrt{1+\varphi^{2}-1}=\varphi$$

).  Now, writing $y_{k}=\varphi-x_{k}$, we have $y_{1}=\varphi-1=1/\varphi$, $0<y_{k}\leq\varphi-1<\varphi^{2}$ and

	$\displaystyle y_{k}$	$\displaystyle=\varphi-\sqrt{1+x_{k-1}}$
		$\displaystyle=\varphi-\sqrt{1+\varphi-y_{k-1}}$
		$\displaystyle=\varphi-\sqrt{\varphi^{2}-y_{k-1}}$
		$\displaystyle<\dfrac{y_{k-1}}{\varphi}<\dfrac{y_{k-2}}{\varphi^{2}}<\dfrac{y_{k-3}}{\varphi^{3}}$

because

$$\begin{array}[c]{ccccc}\dfrac{y_{k-1}}{\varphi}<\varphi,&&\text{i.e.,}&&\dfrac{y_{k-1}}{\varphi^{2}}<1\end{array}$$

hence

$$\begin{array}[c]{ccccc}0<\varphi-\dfrac{y_{k-1}}{\varphi},&&\text{i.e.,}&&\dfrac{y_{k-1}^{2}}{\varphi^{2}}<y_{k-1}\end{array}$$

hence

$$\varphi-\dfrac{y_{k-1}}{\varphi}=\sqrt{\varphi^{2}-2y_{k-1}+\dfrac{y_{k-1}^{2}}{\varphi^{2}}}<\sqrt{\varphi^{2}-y_{k-1}}$$

hence

$$\varphi-\sqrt{\varphi^{2}-y_{k-1}}<\dfrac{y_{k-1}}{\varphi};$$

thus

$$y_{k}<\dfrac{y_{1}}{\varphi^{k-1}}=\frac{1}{\varphi^{k}}$$

for all $k$.

From $x_{k-1}=\varphi-y_{k-1}$, we observe

$$x_{k}^{2}=1+x_{k-1}=1+\varphi-y_{k-1}=\varphi^{2}-y_{k-1}$$

hence

$$y_{k-1}=\varphi^{2}-x_{k}^{2}=(\varphi-x_{k})(\varphi+x_{k})=y_{k}(\varphi+x_{k})$$

hence

$$\frac{2\varphi}{\varphi+x_{k}}=2\varphi\,\frac{y_{k}}{y_{k-1}}$$

hence

$${\displaystyle\prod\limits_{k=2}^{n}}\,\frac{2\varphi}{\varphi+x_{k}}={\displaystyle\prod\limits_{k=2}^{n}}\,2\varphi\,\frac{y_{k}}{y_{k-1}}=(2\varphi)^{n-1\,}\frac{y_{n}}{y_{1}}=\frac{(2\varphi)^{n}}{2}y_{n}$$

hence

$$\frac{\varphi-x_{n}}{2}=\frac{y_{n}}{2}=\frac{1}{(2\varphi)^{n}}\,{\displaystyle\prod\limits_{k=2}^{n}}\,\frac{2\varphi}{\varphi+x_{k}}$$

because $y_{1}=1/\varphi$; therefore

$$C=\lim_{n\rightarrow\infty}\,(2\varphi)^{n}\,\frac{\varphi-x_{n}}{2}={\displaystyle\prod\limits_{k=2}^{\infty}}\,\frac{2\varphi}{\varphi+x_{k}}$$

exists and is nonzero since

$${\displaystyle\sum\limits_{k=2}^{\infty}}\left(\frac{2\varphi}{\varphi+x_{k}}-1\right)={\displaystyle\sum\limits_{k=2}^{\infty}}\,\frac{\varphi-x_{k}}{\varphi+x_{k}}<{\displaystyle\sum\limits_{k=2}^{\infty}}\,y_{k}<{\displaystyle\sum\limits_{k=2}^{\infty}}\,\frac{1}{\varphi^{k}}$$

converges.  This completes the proof.  The numerically-efficient expression for

$$C=1.0986419643941564857346689...=\frac{1}{2}\left(2.1972839287883129714693378...\right)$$

as an infinite product did not appear in [4], but was subsequently found by Philippe Flajolet & Paul Zimmermann several years later [5].  No trace of their derivation has survived; the above proof is new.  Other calculations of $C$ are exhibited in [6, 7].

In contrast, an analysis of the recurrence

$x_{1}=0,$ $x_{k}=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}x_{k-1}}$ for $k\geq 2$

is trivial [8]:

$$\begin{array}[c]{ccccc}x_{k}=\cos\left(\dfrac{\pi}{2^{k}}\right),&&x_{k}\rightarrow 1^{-},&&\lim\limits_{k\rightarrow\infty}\,4^{k}(1-x_{k})=\dfrac{\pi^{2}}{2}.\end{array}$$

Finally, the recurrence

$x_{1}=1,$ $x_{k}=\sqrt{\,2+2x_{k-1}}$ for $k\geq 2$

approaches $1+\sqrt{3}$, but higher-order asymptotics (akin to [4]) await discovery.

Given the recurrence

$$\begin{array}[c]{ccccc}x_{0}=1,&&x_{k}=\dfrac{1+\sqrt{4x_{k-1}^{2}+1}}{2}&&\text{for }k\geq 1\end{array}$$

we start with asymptotics [9]

$$x_{k}\sim\frac{1}{2}k+\frac{1}{4}\ln(k)+C$$

valid as $k\rightarrow\infty$, for some constant $C=C(x_{0})$.  On the basis of numerical experimentation, we hypothesize that the next terms of the asymptotic series must be of the form

	$\displaystyle p_{1}\,\frac{\ln(k)}{k}+\frac{p_{0}}{k}+q_{2}\,\frac{\ln(k)^{2}}{k^{2}}+q_{1}\,\frac{\ln(k)}{k^{2}}+\frac{q_{0}}{k^{2}}+r_{3}\,\frac{\ln(k)^{3}}{k^{3}}+r_{2}\,\frac{\ln(k)^{2}}{k^{3}}$
	$\displaystyle+r_{1}\,\frac{\ln(k)}{k^{3}}+\frac{r_{0}}{k^{3}}+s_{4}\,\frac{\ln(k)^{4}}{k^{4}}+s_{3}\,\frac{\ln(k)^{3}}{k^{4}}+s_{2}\,\frac{\ln(k)^{2}}{k^{4}}+s_{1}\,\frac{\ln(k)}{k^{4}}+\frac{s_{0}}{k^{4}}.$

The challenge is to express each coefficient $p_{j}$, $q_{j}$, $r_{j}$, $s_{j}$ as a polynomial in $C$.  To find these, we replace $k$ by $k+1$ everywhere:

$$\frac{1}{2}(k+1)+\frac{1}{4}\ln(k+1)+C+p_{1}\,\frac{\ln(k+1)}{k+1}+\cdots+\frac{s_{0}}{(k+1)^{4}}$$

and expand in powers of $k$ and $\ln(k)$:

$$\ln(k+1)\sim\frac{1}{k}-\frac{1}{2k^{2}}+\frac{1}{3k^{3}}-\frac{1}{4k^{4}}+\frac{1}{5k^{5}}-+\cdots,$$

$$\frac{\ln(k+1)}{k+1}=\left(\frac{1}{k}-\frac{1}{k^{2}}+\frac{1}{k^{3}}-\frac{1}{k^{4}}+\frac{1}{k^{5}}-+\cdots\right)\ln(k)+\left(\frac{1}{k^{2}}-\frac{3}{2k^{3}}+\frac{11}{6k^{4}}-\frac{25}{12k^{5}}+-\cdots\right),$$

$$\frac{1}{k+1}\sim\frac{1}{k}-\frac{1}{k^{2}}+\frac{1}{k^{3}}-\frac{1}{k^{4}}+\frac{1}{k^{5}}-+\cdots,$$

$$\frac{\ln(k+1)^{2}}{(k+1)^{2}}\sim\left(\frac{1}{k^{2}}-\frac{2}{k^{3}}+\frac{3}{k^{4}}-\frac{4}{k^{5}}+\cdots\right)\ln(k)^{2}+\left(\frac{2}{k^{3}}-\frac{5}{k^{4}}+\frac{26}{3k^{5}}-\cdots\right)\ln(k)+\left(\frac{1}{k^{4}}-\frac{3}{k^{5}}+\cdots\right),$$

$$\frac{\ln(k+1)}{(k+1)^{2}}\sim\left(\frac{1}{k^{2}}-\frac{2}{k^{3}}+\frac{3}{k^{4}}-\frac{4}{k^{5}}+-\cdots\right)\ln(k)+\left(\frac{1}{k^{3}}-\frac{5}{2k^{4}}+\frac{13}{3k^{5}}-+\cdots\right),$$

$$\frac{1}{(k+1)^{2}}\sim\frac{1}{k^{2}}-\frac{2}{k^{3}}+\frac{3}{k^{4}}-\frac{4}{k^{5}}+-\cdots,$$

$$\frac{\ln(k+1)^{3}}{(k+1)^{3}}\sim\left(\frac{1}{k^{3}}-\frac{3}{k^{4}}+\frac{6}{k^{5}}-+\cdots\right)\ln(k)^{3}+\left(\frac{3}{k^{4}}-\frac{21}{2k^{5}}-+\cdots\right)\ln(k)^{2}+\left(\frac{3}{k^{5}}+-\cdots\right)\ln(k),$$

$$\frac{\ln(k+1)^{2}}{(k+1)^{3}}\sim\left(\frac{1}{k^{3}}-\frac{3}{k^{4}}+\frac{6}{k^{5}}-+\cdots\right)\ln(k)^{2}+\left(\frac{2}{k^{4}}-\frac{7}{k^{5}}+-\cdots\right)\ln(k)+\left(\frac{1}{k^{5}}-+\cdots\right),$$

$$\frac{\ln(k+1)}{(k+1)^{3}}\sim\left(\frac{1}{k^{3}}-\frac{3}{k^{4}}+\frac{6}{k^{5}}-+\cdots\right)\ln(k)+\left(\frac{1}{k^{4}}-\frac{7}{2k^{5}}+-\cdots\right),$$

$$\frac{1}{(k+1)^{3}}\sim\frac{1}{k^{3}}-\frac{3}{k^{4}}+\frac{6}{k^{5}}-+\cdots,$$

$$\frac{\ln(k+1)^{4}}{(k+1)^{4}}\sim\left(\frac{1}{k^{4}}-\frac{4}{k^{5}}+-\cdots\right)\ln(k)^{4}+\left(\frac{4}{k^{5}}-+\cdots\right)\ln(k)^{3},$$

$$\frac{\ln(k+1)^{3}}{(k+1)^{4}}\sim\left(\frac{1}{k^{4}}-\frac{4}{k^{5}}+-\cdots\right)\ln(k)^{3}+\left(\frac{3}{k^{5}}-+\cdots\right)\ln(k)^{2},$$

$$\frac{\ln(k+1)^{2}}{(k+1)^{4}}\sim\left(\frac{1}{k^{4}}-\frac{4}{k^{5}}+-\cdots\right)\ln(k)^{2}+\left(\frac{2}{k^{5}}-+\cdots\right)\ln(k),$$

$$\frac{\ln(k+1)}{(k+1)^{4}}\sim\left(\frac{1}{k^{4}}-\frac{4}{k^{5}}+-\cdots\right)\ln(k)+\left(\frac{1}{k^{5}}-+\cdots\right),$$

$$\frac{1}{(k+1)^{4}}\sim\frac{1}{k^{4}}-\frac{4}{k^{5}}+-\cdots.$$

To avoid dealing with the radical in

$$2x_{k+1}-1=\sqrt{4x_{k}^{2}+1}$$

square both sides, obtaining

$$4x_{k+1}^{2}-4x_{k+1}+1=4x_{k}^{2}+1$$

i.e.,

$$x_{k+1}^{2}-x_{k+1}=x_{k}^{2}.$$

Upon rearrangement, relevant terms (in decreasing order of significance) of $x_{k+1}^{2}-x_{k+1}$ become

	$\displaystyle\left(\frac{1}{8}-p_{1}+\frac{p_{0}}{2}+2p_{1}C+q_{1}\right)\frac{\ln(k)}{k}+\left(-\frac{1}{8}+p_{1}-p_{0}+\frac{C}{2}+2p_{0}C+q_{0}\right)\frac{1}{k}$
	$\displaystyle+\left(-\frac{p_{1}}{2}+p_{1}^{2}+r_{2}-2q_{2}+2q_{2}C+\frac{q_{1}}{2}\right)\frac{\ln(k)^{2}}{k^{2}}$
	$\displaystyle+\left(-\frac{1}{16}+2p_{1}-\frac{p_{0}}{2}+2p_{1}p_{0}-2p_{1}C+r_{1}+2q_{2}-2q_{1}+2q_{1}C+\frac{q_{0}}{2}\right)\frac{\ln(k)}{k^{2}}$
	$\displaystyle+\left(\frac{7}{48}-\frac{3p_{1}}{2}+\frac{3p_{0}}{2}+p_{0}^{2}-\frac{C}{4}+2p_{1}C-2p_{0}C+r_{0}+q_{1}-2q_{0}+2q_{0}C\right)\frac{1}{k^{2}}$
	$\displaystyle+\left(s_{3}-3r_{3}+2r_{3}C+\frac{r_{2}}{2}-q_{2}+2p_{1}q_{2}\right)\frac{\ln(k)^{3}}{k^{3}}$
	$\displaystyle+\left(\frac{p_{1}}{2}-2p_{1}^{2}+s_{2}+3r_{3}-3r_{2}+2r_{2}C+\frac{r_{1}}{2}+\frac{9q_{2}}{2}+2p_{0}q_{2}-4q_{2}C-q_{1}+2p_{1}q_{1}\right)\frac{\ln(k)^{2}}{k^{3}}$
	$\displaystyle+\left(\frac{1}{24}-\frac{5p_{1}}{2}+2p_{1}^{2}+\frac{p_{0}}{2}-4p_{1}p_{0}+2p_{1}C+s_{1}+2r_{2}-3r_{1}+2r_{1}C+\frac{r_{0}}{2}-5q_{2}\right.$
	$\displaystyle\left.\;\;\;\;\;+4q_{2}C+4q_{1}+2p_{0}q_{1}-4q_{1}C-q_{0}+2p_{1}q_{0}\right)\frac{\ln(k)}{k^{3}}$

	$\displaystyle+\left(-\frac{1}{8}+\frac{7p_{1}}{3}-\frac{7p_{0}}{4}+2p_{1}p_{0}-2p_{0}^{2}+\frac{C}{6}-3p_{1}C+2p_{0}C+s_{0}+r_{1}-3r_{0}\right.$
	$\displaystyle\left.\;\;\;\;\;+2r_{0}C+q_{2}-\frac{5q_{1}}{2}+2q_{1}C+\frac{7q_{0}}{2}+2p_{0}q_{0}-4q_{0}C\right)\frac{1}{k^{3}}$
	$\displaystyle+\left(-4s_{4}+2s_{4}C+\frac{s_{3}}{2}-\frac{3r_{3}}{2}+2p_{1}r_{3}+q_{2}^{2}\right)\frac{\ln(k)^{4}}{k^{4}}$
	$\displaystyle+\left(4s_{4}-4s_{3}+2s_{3}C+\frac{s_{2}}{2}+8r_{3}+2p_{0}r_{3}-6r_{3}C-\frac{3r_{2}}{2}+2p_{1}r_{2}+\frac{3q_{2}}{2}-6p_{1}q_{2}+2q_{2}q_{1}\right)\frac{\ln(k)^{3}}{k^{4}}$
	$\displaystyle+\left(-\frac{p_{1}}{2}+3p_{1}^{2}+3s_{3}-4s_{2}+2s_{2}C+\frac{s_{1}}{2}-\frac{21r_{3}}{2}+6r_{3}C+\frac{15r_{2}}{2}+2p_{0}r_{2}-6r_{2}C-\frac{3r_{1}}{2}\right.$
	$\displaystyle\left.\;\;\;\;\;+2p_{1}r_{1}-\frac{31q_{2}}{4}+6p_{1}q_{2}-6p_{0}q_{2}+6q_{2}C+\frac{3q_{1}}{2}-6p_{1}q_{1}+q_{1}^{2}+2q_{2}q_{0}\right)\frac{\ln(k)^{2}}{k^{4}}$

	$\displaystyle+\left(-\frac{1}{32}+\frac{17p_{1}}{6}-5p_{1}^{2}-\frac{p_{0}}{2}+6p_{1}p_{0}-2p_{1}C+2s_{2}-4s_{1}+2s_{1}C+\frac{s_{0}}{2}+3r_{3}-7r_{2}+4r_{2}C\right.$
	$\displaystyle\left.\;\;\;\;\;+7r_{1}+2p_{0}r_{1}-6r_{1}C-\frac{3r_{0}}{2}+2p_{1}r_{0}+\frac{61q_{2}}{6}+4p_{0}q_{2}-10q_{2}C-\frac{13q_{1}}{2}\right.$
	$\displaystyle\left.\;\;\;\;\;+4p_{1}q_{1}-6p_{0}q_{1}+6q_{1}C+\frac{3q_{0}}{2}-6p_{1}q_{0}+2q_{1}q_{0}\right)\frac{\ln(k)}{k^{4}}$
	$\displaystyle+\left(\frac{103}{960}-\frac{37p_{1}}{12}+p_{1}^{2}+\frac{23p_{0}}{12}-5p_{1}p_{0}+3p_{0}^{2}-\frac{C}{8}+\frac{11p_{1}C}{3}-2p_{0}C+s_{1}-4s_{0}+2s_{0}C\right.$
	$\displaystyle\left.\;\;\;\;\;+r_{2}-\frac{7r_{1}}{2}+2r_{1}C+\frac{13r_{0}}{2}+2p_{0}r_{0}-6r_{0}C-3q_{2}+2q_{2}C+\frac{29q_{1}}{6}+2p_{0}q_{1}\right.$
	$\displaystyle\left.\;\;\;\;\;-5q_{1}C-\frac{21q_{0}}{4}+2p_{1}q_{0}-6p_{0}q_{0}+6q_{0}C+q_{0}^{2}\right)\frac{1}{k^{4}}.$

Performing an analogous substitution in $x_{k}$, the corresponding terms of $x_{k}^{2}$ become

	$\displaystyle\left(\frac{p_{0}}{2}+2p_{1}C+q_{1}\right)\frac{\ln(k)}{k}+\left(2p_{0}C+q_{0}\right)\frac{1}{k}+\left(p_{1}^{2}+r_{2}+2q_{2}C+\frac{q_{1}}{2}\right)\frac{\ln(k)^{2}}{k^{2}}$
	$\displaystyle+\left(2p_{1}p_{0}+r_{1}+2q_{1}C+\frac{q_{0}}{2}\right)\frac{\ln(k)}{k^{2}}+\left(p_{0}^{2}+r_{0}+2q_{0}C\right)\frac{1}{k^{2}}$
	$\displaystyle+\left(s_{3}+2r_{3}C+\frac{r_{2}}{2}+2p_{1}q_{2}\right)\frac{\ln(k)^{3}}{k^{3}}+\left(s_{2}+2r_{2}C+\frac{r_{1}}{2}+2p_{0}q_{2}+2p_{1}q_{1}\right)\frac{\ln(k)^{2}}{k^{3}}$
	$\displaystyle+\left(s_{1}+2r_{1}C+\frac{r_{0}}{2}+2p_{0}q_{1}+2p_{1}q_{0}\right)\frac{\ln(k)}{k^{3}}+\left(s_{0}+2r_{0}C+2p_{0}q_{0}\right)\frac{1}{k^{3}}$

	$\displaystyle+\left(2s_{4}C+\frac{s_{3}}{2}+2p_{1}r_{3}+q_{2}^{2}\right)\frac{\ln(k)^{4}}{k^{4}}+\left(2s_{3}C+\frac{s_{2}}{2}+2p_{0}r_{3}+2p_{1}r_{2}+2q_{2}q_{1}\right)\frac{\ln(k)^{3}}{k^{4}}$
	$\displaystyle+\left(2s_{2}C+\frac{s_{1}}{2}+2p_{0}r_{2}+2p_{1}r_{1}+q_{1}^{2}+2q_{2}q_{0}\right)\frac{\ln(k)^{2}}{k^{4}}$
	$\displaystyle+\left(2s_{1}C+\frac{s_{0}}{2}+2p_{0}r_{1}+2p_{1}r_{0}+2q_{1}q_{0}\right)\frac{\ln(k)}{k^{4}}+\left(2s_{0}C+2p_{0}r_{0}+q_{0}^{2}\right)\frac{1}{k^{4}}.$

Matching coefficients, we obtain

$$\begin{array}[c]{ccc}p_{1}=\dfrac{1}{8},&&p_{0}=\dfrac{1}{2}C\end{array}$$

which are consistent with [10] and

$$\begin{array}[c]{ccccc}q_{2}=-\dfrac{1}{32},&&q_{1}=-\left(\dfrac{1}{4}C-\dfrac{1}{16}\right),&&q_{0}=-\left(\dfrac{1}{2}C^{2}-\dfrac{1}{4}C-\dfrac{1}{96}\right),\end{array}$$

$$\begin{array}[c]{ccccc}r_{3}=\dfrac{1}{96},&&r_{2}=\dfrac{1}{8}C-\dfrac{3}{64},&&r_{1}=\dfrac{1}{2}C^{2}-\dfrac{3}{8}C+\dfrac{1}{48},\end{array}$$

$$r_{0}=\frac{2}{3}C^{3}-\dfrac{3}{4}C^{2}+\dfrac{1}{12}C+\dfrac{7}{576},$$

$$\begin{array}[c]{ccccc}s_{4}=-\dfrac{1}{256},&&s_{3}=-\left(\dfrac{1}{16}C-\dfrac{11}{384}\right),&&s_{2}=-\left(\dfrac{3}{8}C^{2}-\dfrac{11}{32}C+\dfrac{5}{128}\right),\end{array}$$

$$\begin{array}[c]{ccc}s_{1}=-\left(C^{3}-\dfrac{11}{8}C^{2}+\dfrac{5}{16}C+\dfrac{1}{128}\right),&&s_{0}=-\left(C^{4}-\dfrac{11}{6}C^{3}+\dfrac{5}{8}C^{2}+\dfrac{1}{32}C-\dfrac{47}{5760}\right)\end{array}$$

which are new (as far as is known).

These fourteen parameter values allow us to estimate the constant $C$.  Our simple procedure involves computing $a_{10000000000}$ exactly via recursion, setting this equal to our series (up to $s_{0}/k^{4}$) and then solving:

$$C=0.8232354508791921603541165....$$

Note that the estimate $2\,C\approx 1.6464707$ appears in [11].  We find the implicit representation

$$x_{k+1}(x_{k+1}-1)=x_{k}^{2}$$

to be intriguing: the left-hand side echoes the logistic map $\xi\,(1-\xi)$ but only somewhat: it is off by a sign.  In the following section, a comparable implicit representation leads to a surprising outcome.

Given the recurrence

$$\begin{array}[c]{ccccc}x_{0}=0,&&x_{k}=\dfrac{x_{k-1}+\sqrt{x_{k-1}^{2}+4}}{2}&&\text{for }k\geq 1\end{array}$$

we start with conjectured asymptotics (with no known theoretical basis)

$$x_{k}\sim\sqrt{2k}-\frac{1}{4\sqrt{2}}\frac{\ln(k)}{k^{1/2}}-\frac{C}{k^{1/2}}$$

as $k\rightarrow\infty$, for some constant $C=C(x_{0})$.  It is reasonable to hypothesize that the next terms of this series are

$$p_{2}\,\frac{\ln(k)^{2}}{k^{3/2}}+p_{1}\,\frac{\ln(k)}{k^{3/2}}+\frac{p_{0}}{k^{3/2}}+q_{3}\,\frac{\ln(k)^{3}}{k^{5/2}}+q_{2}\,\frac{\ln(k)^{2}}{k^{5/2}}+q_{1}\,\frac{\ln(k)}{k^{5/2}}+\frac{q_{0}}{k^{5/2}}.$$

To find $p_{j}$, $q_{j}$ we replace $k$ by $k+1$ everywhere:

$$\sqrt{2(k+1)}-\frac{1}{4\sqrt{2}}\frac{\ln(k+1)}{(k+1)^{1/2}}+\frac{C}{(k+1)^{1/2}}+p_{2}\,\frac{\ln(k+1)^{2}}{(k+1)^{3/2}}+\cdots+\frac{q_{0}}{(k+1)^{5/2}}$$

and expand in powers of $k$ and $\ln(k)$:

$$(k+1)^{1/2}\sim k^{1/2}+\frac{1}{2k^{1/2}}-\frac{1}{8k^{3/2}}+\frac{1}{16k^{5/2}}-\frac{5}{128k^{7/2}}+\frac{7}{256k^{9/2}}-+\cdots,$$

	$\displaystyle\frac{\ln(k+1)}{(k+1)^{1/2}}$	$\displaystyle\sim\left(\frac{1}{k^{1/2}}-\frac{1}{2k^{3/2}}+\frac{3}{8k^{5/2}}-\frac{5}{16k^{7/2}}+\frac{35}{128k^{9/2}}-+\cdots\right)\ln(k)$
		$\displaystyle+\left(\frac{1}{k^{3/2}}-\frac{1}{k^{5/2}}+\frac{23}{24k^{7/2}}-\frac{11}{12k^{9/2}}+-\cdots\right),$

$$\frac{1}{(k+1)^{1/2}}\sim\frac{1}{k^{1/2}}-\frac{1}{2k^{3/2}}+\frac{3}{8k^{5/2}}-\frac{5}{16k^{7/2}}+\frac{35}{128k^{9/2}}-+\cdots,$$

	$\displaystyle\frac{\ln(k+1)^{2}}{(k+1)^{3/2}}$	$\displaystyle\sim\left(\frac{1}{k^{3/2}}-\frac{3}{2k^{5/2}}+\frac{15}{8k^{7/2}}-\frac{35}{16k^{9/2}}+-\cdots\right)\ln(k)^{2}$
		$\displaystyle+\left(\frac{2}{k^{5/2}}-\frac{4}{k^{7/2}}+\frac{71}{12k^{9/2}}-+\cdots\right)\ln(k)+\left(\frac{1}{k^{7/2}}-\frac{5}{2k^{9/2}}+-\cdots\right),$

$$\frac{\ln(k+1)}{(k+1)^{3/2}}\sim\left(\frac{1}{k^{3/2}}-\frac{3}{2k^{5/2}}+\frac{15}{8k^{7/2}}-\frac{35}{16k^{9/2}}+\cdots\right)\ln(k)+\left(\frac{1}{k^{5/2}}-\frac{2}{k^{7/2}}+\frac{71}{24k^{9/2}}-\cdots\right),$$

$$\frac{1}{(k+1)^{3/2}}\sim\frac{1}{k^{3/2}}-\frac{3}{2k^{5/2}}+\frac{15}{8k^{7/2}}-\frac{35}{16k^{9/2}}+-\cdots,$$

	$\displaystyle\frac{\ln(k+1)^{3}}{(k+1)^{5/2}}$	$\displaystyle\sim\left(\frac{1}{k^{5/2}}-\frac{5}{2k^{7/2}}+\frac{35}{8k^{9/2}}-+\cdots\right)\ln(k)^{3}$
		$\displaystyle+\left(\frac{3}{k^{7/2}}-\frac{9}{k^{9/2}}+-\cdots\right)\ln(k)^{2}+\left(\frac{3}{k^{9/2}}-+\cdots\right)\ln(k),$

$$\frac{\ln(k+1)^{2}}{(k+1)^{5/2}}\sim\left(\frac{1}{k^{5/2}}-\frac{5}{2k^{7/2}}+\frac{35}{8k^{9/2}}-\cdots\right)\ln(k)^{2}+\left(\frac{2}{k^{7/2}}-\frac{6}{k^{9/2}}+\cdots\right)\ln(k)+\left(\frac{1}{k^{9/2}}-\cdots\right),$$

$$\frac{\ln(k+1)}{(k+1)^{5/2}}\sim\left(\frac{1}{k^{5/2}}-\frac{5}{2k^{7/2}}+\frac{35}{8k^{9/2}}-+\cdots\right)\ln(k)+\left(\frac{1}{k^{7/2}}-\frac{3}{k^{9/2}}+-\cdots\right),$$

$$\frac{1}{(k+1)^{5/2}}\sim\frac{1}{k^{5/2}}-\frac{5}{2k^{7/2}}+\frac{35}{8k^{9/2}}-+\cdots.$$

To avoid dealing with the radical in

$$2x_{k+1}-x_{k}=\sqrt{x_{k}^{2}+4}$$

square both sides, obtaining

$$4x_{k+1}^{2}-4x_{k+1}x_{k}+x_{k}^{2}=x_{k}^{2}+4$$

i.e.,

$$x_{k+1}^{2}-1=x_{k+1}x_{k}$$

i.e.,

$$x_{k+1}-\frac{1}{x_{k+1}}=x_{k}.$$

Upon rearrangement, the terms of $x_{k+1}-1/x_{k+1}$ involving either $k^{5/2}$ or $k^{7/2}$ become most relevant.  Performing an analogous substitution in $x_{k}$, we match coefficients as before.  The terms containing $k^{5/2}$ give

$$\begin{array}[c]{ccccc}p_{2}=-\dfrac{1}{64\sqrt{2}},&&p_{1}=-\dfrac{4C-\sqrt{2}}{32},&&p_{0}=-\dfrac{8\sqrt{2}C^{2}-8C+\sqrt{2}}{32}\end{array}$$

and the terms containing $k^{7/2}$ give

$$\begin{array}[c]{ccccc}q_{3}=-\dfrac{1}{512\sqrt{2}},&&q_{2}=-\dfrac{3C-\sqrt{2}}{128},&&q_{1}=-\dfrac{24\sqrt{2}C^{2}-32C+5\sqrt{2}}{256},\end{array}$$

$$q_{0}=-\frac{192C^{3}-192\sqrt{2}C^{2}+120C-11\sqrt{2}}{768}.$$

It is astonishing that we have seen these coefficients before – review our analysis [2] of the recurrence

$$\xi_{k+1}=\xi_{k}+\frac{1}{\xi_{k}}$$

which yields the identical seven expressions – although the signs preceding each coefficient may differ.  This is very surprising!  While the sequences behave distinctly, there is a hidden commonality in structure, captured by the polynomials in $C$.