Limit of $x_n^3/n^2$ when $x_{n+1}=x_n+ 1/\sqrt {x_n}$ with $x_0 \gt 0$

Question

Let $(x_n)_{n \ge 0}$ a sequence of real numbers with $x_0 \gt 0$ and $x_{n+1}=x_n+ \frac {1}{\sqrt {x_n}}$.

Check the existence and find $$L=\lim_{n \rightarrow \infty} \frac {x_n^3} {n^2}$$

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Because $x_n$ is increasing, there is $\lim_{n \rightarrow \infty} {x_n}=l$ and cannot be finite (otherwise $l=l + \frac 1 {\sqrt {l}}$ impossible) therefore $L$ is indeterminate of $\frac {\infty}{\infty}$ form.

Answer 1

Using Stolz theorem one can get $$ L = \lim_{n\to \infty} \frac{x_n^3}{n^2} = \lim_{n\to \infty}\frac{x_{n+1}^3-x_n^3}{2n+1} = \lim_{n\to \infty} \frac{3x_n^{3/2} + 3 + \frac{1}{x_n^{3/2}}}{2n+1} = \lim_{n\to \infty} \frac{3x_n^{3/2}}{2n+1} $$ Notice that $$ \frac{3x_n^{3/2}}{2n+1} \sim \frac{3x_n^{3/2}}{2n} $$ That's why we get $$ L = \frac{3}{2}\sqrt{L} $$ and $L=0$ or $L=9/4$ or $L=\infty$ or the limit does not exist. Let's prove that the case $L=0$ is impossible. By induction we will prove $x_n^3 \ge n^2$. The base of induction is obvious. The induction step $$ x_{n+1}^3 = x_n^3 + 3x_n^{3/2} + 3 + \frac{1}{x_n^{3/2}} \ge n^2 + 3n + 3 \ge (n+1)^2 $$ So the case $L=0$ is impossible. Similarly one can show that $L\neq \infty$. So $L = 9/4$ if someone will prove that it exists. Currently I don't know if it exists for any $x_0$ or not.

Answer 2

This was my first attempt to avoid Stolz-Cesàro, just for fun. The second failed.

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The limit exists and is finite, due to $\dfrac{x_n^3}{n^2}$ being decreasing (and obviously positive). Indeed, for $n\ge1$, we have \begin{align}\frac{x_{n+1}^3}{(n+1)^2}&< \frac{x_n^3}{n^2} \\ \left(1+x_n^{-3/2}\right)^3&< 1+\frac2n+\frac1{n^2} \\ 3x_n^{-3/2}+3x_n^{-3}+x_n^{-9/2}&<\frac2n+\frac1{n^2},\end{align}and after rearranging and using Cardano-Tartaglia's formula (of course I exploited Wolfram Alpha) we get $$x_n^{3/2}>\frac{n^2}{2n+1}+\frac{a_n}{\sqrt[3]{2}(2n+1)}+\frac{\sqrt[3]{2}n^2(n+1)^2}{a_n(2n+1)},\tag{$\star$}$$where $a_n=\sqrt[3]{2n^6+6n^5+7n^4+4n^3+n^2+\sqrt{4n^{10}+20n^9+41n^8+44n^7+26n^6+8n^5+n^4}}.$ We need to strictly bound $a_n$ from above and below in order to strengthen $(\star)$ and prove the resulting inequality by induction. First of all, the square root inside the cube root, let's call it $b_n$, satisfies $$\require{color} 1+\frac5{2n}<\sqrt{1+\frac5n+\frac{10}{n^2}}<\frac{b_n}{2n^5}<\sqrt{1+\frac5n+\frac{30}{n^2}}<1+\frac5{2n}+\frac{\color\red{12}}{n^2} $$ hence $$\require{color} \begin{align}\sqrt[3]{2n^6+8n^5+12n^4+4n^3+n^2}&< \ a_n<\sqrt[3]{2n^6+8n^5+12n^4+28n^3+n^2} \\ \sqrt[3]{1+\frac4n+\frac6{n^2}+\frac2{n^3}}&<\frac{a_n}{\sqrt[3]{2}n^2}<\sqrt[3]{1+\frac4n+\frac6{n^2}+\frac{15}{n^3}}\\ 1+\frac4{3n}+\frac{\color\red{1/56}}{n^2} &<\frac{a_n}{\sqrt[3]{2}n^2}<1+\frac4{3n}+\frac{\color\red{2/9}}{n^2}, \end{align}$$ where the red coefficients can be found by setting, for positive $A, B, C$ respectively, $$1+\frac5n+\frac{30}{n^2}<\left(1+\frac5{2n}+\frac{A}{n^2}\right)^2,$$ $$1+\frac4n+\frac6{n^2}+\frac2{n^3}>\left(1+\frac4{3n}+\frac{B}{n^2}\right)^3 $$ and $$1+\frac4n+\frac6{n^2}+\frac{15}{n^3}<\left(1+\frac4{3n}+\frac{C}{n^2}\right)^3;$$ due to the direction of the inequalities and the positiveness of all the terms, $A=12$ and $C=2/9$ are easy, as it's enough to equate the resulting coefficients of the $n^{-2}$ terms. As for $B$, looking at the $n^{-2}$ terms is not sufficient but does tell us $B<2/9,$ therefore, the inequality, which once rearranged and simplified is $$\frac23-3B>\frac{8B+10/27}{n}+\frac{B^2+\frac{16}{3}B}{n^2}+\frac{\frac{4}{3}B^2}{n^3}+\frac{ B^3}{n^4},$$ is weaker than $69B^2+441B-8<0,$ i.e., $B$ $\require{color} \color\red{<\dfrac1{56}<\dfrac{16}{885}}$$\require{color} \color\black{<\dfrac{16}{441+\sqrt{196689}}}.$

Then, the following inequality is tighter than $(\star)$: \begin{align}x_n^{3/2}(2n+1)>n^2+n^2\left(1+\frac4{3n}+\frac2{9n^2}\right)+(n+1)^2\left(1+\frac4{3n}+\frac1{56n^2}\right)^{-1};\end{align} similarly as above, one can find $$(n+1)^2\left(1+\frac4{3n}+\frac1{56n^2}\right)^{-1}<n^2+\frac23n+\frac{1}{10}, $$so after rearrangement and some last trivial estimates, we conclude that $(\star)$ follows from $ x_n^{3/2}>\dfrac32n+1,$ or equivalently, $x_n^3>\dfrac94n^2+3n+1.$ As announced, we'll prove this by induction: firstly,\begin{align} \require{color}x_1^{3}=\left(x_0+\frac1{\sqrt{x_0}}\right)^{3}\color\red{\ge}\frac{27}{4}&>\frac{25}4,\end{align} $\require{color}\color\red{\text{because $f(x)=x+x^{-1/2}$ has an absolute minimum, that is $3\cdot2^{-2/3}$}}$;

finally, the inductive step is\begin{align}x_{k+1}^3=x_k^3+3x_k^{3/2}+3+x_k^{-3/2}&> \frac94k^2+3k+1+\frac92k+3+3 \\ &=\frac94k^2+\frac{15}2k+7 \\ &>\frac94k^2+\frac{15}2k+\frac{25}4=\frac94(k+1)^2+3(k+1)+1.\end{align} Do note that $(\star)$ implies $L\ge9/4$.

So $x_n^3=L \ n^2(1+o(1))$, whence $x_n^{3/2}= \sqrt{L} \ n +u_n$ and $x_n=L^{1/3}n^{2/3}+v_n,$ with $u_n=o(n), v_n=o(n^{2/3}).$

Proposition. $v_{n+1}-v_n=o(n^{-1/3})$.

Proof. Assume there exists a constant $c>0$ such that, for all large enough $n$, $n^{1/3}\left(v_{n+1}-v_n\right)>c$. The inequality can be rewritten as \begin{align} n^{1/3}\left(x_{n+1}-x_n-L^{1/3}\left[(n+1)^{2/3}-n^{2/3}\right]\right)&>c \\ \frac{n^{1/3}}{\sqrt{x_n}}-L^{1/3}n\left[\left(1+\frac1n\right)^{2/3}-1\right] &> c,\end{align} and since the LHS converges to $L^{-1/6}-\frac{2}{3}L^{1/3},$ we must have $L^{-1/6}-\frac{2}{3}L^{1/3}\ge c,$ which amounts to another cubic inequality and gives$$\require{color}L\le\left(\frac12\left(\sqrt[3]{c^3+3\sqrt{2c^3+9}+9}+\dfrac{c^2}{\sqrt[3]{c^3+3\sqrt{2c^3+9}+9}}\right)-c\right)^3\color\red{<((3/2)^{2/3})^3}=\frac94,$$ $\require{color}\color\red{\text{by the decreasingness of that function on $\mathbb{R^+}$}}$, contradicting $(\star).$ We're done.$\qquad \ \ \ \ \ \ \ \ \ \ \ \ $ $\square$

Now, we have $$\dfrac{x_{n+1}}{x_n}=1+x_n^{-3/2},$$ therefore, \begin{align}1=\lim_{n\to\infty} x_n^{3/2}\log\left(\frac{x_{n+1}}{x_n}\right)&=\lim_{n\to\infty}\sqrt{L}\ n\log\left(\frac{L^{1/3}(n+1)^{2/3}+v_{n+1}}{L^{1/3}n^{2/3}+v_n}\right) \\ &=\lim_{n\to\infty}\sqrt{L}\ n\left(\frac{L^{1/3}(n+1)^{2/3}+v_{n+1}}{L^{1/3}n^{2/3}+v_n}-1\right) \\ &=\lim_{n\to\infty}\sqrt{L}\ n\frac{L^{1/3}\left[(n+1)^{2/3}-n^{2/3}\right]+v_{n+1}-v_n}{L^{1/3}n^{2/3}+v_n}\\&=\lim_{n\to\infty}\sqrt{L}\ n\left[\left(1+\frac1n\right)^{2/3}-1\right]+\sqrt{L}\ n\frac{v_{n+1}-v_n}{L^{1/3}n^{2/3}}\\ &= \frac23 \sqrt{L}+\lim_{n\to\infty}L^{1/6}n^{1/3}\left(v_{n+1}-v_n\right).\end{align}By the proposition above, the last limit is $0$, hence we conclude $L=9/4.$

Answer 3

We will use Stolz theorem. Many times before applying this theorem it is a good idea to make the denominator linear.

We do so by observing we can calculate $\sqrt L= \lim _n \frac{x_n^{3/2}}{n}$ instead.

\begin{align*} \lim _n \frac{x_n^{3/2}}{n} &=^{\text{ST}} \lim _n x_{n+1}^{3/2}- x_n^{3/2} \\ &= \lim _n \frac{x_{n+1}^{3}- x_n^{3}}{ x_{n+1}^{3/2}+ x_n^{3/2}}\\ &= \lim _n \frac{3 x_n^{3/2}}{ x_{n+1}^{3/2}+ x_n^{3/2}}\text{, see Vortuoz's answer}\\ &= \lim _n \frac{3 x_n^{3/2}}{ ( x_n +\frac{1}{\sqrt{x_n}} )^{3/2}+ x_n^{3/2}} \\ &= \lim _n \frac{3 }{ ( 1 +\frac{1}{\sqrt{x_n^3}} )^{3/2}+ 1} \\ &=\frac{3}{2} \end{align*} In the second to last line we used that $\lim _n x_n=\infty$.

Therefore $$L=\frac{9}{4}.$$

Answer 4

Just for reference, we can prove that $(x_n)$ has the following asymptotic expansion: There exists a constant $C$, depending only on $x_0$, such that

$$ x_n^{3/2} = \frac{3}{2}n + \frac{1}{4}\log n + C + \mathcal{O}\left( \frac{\log n}{n} \right) \tag{1} $$

where the implicit bound for $\mathcal{O}$ possibly depends on $x_0$. Here are some explanations:

1. Continuum analogue of the problem $y'(t) = y(t)^{-1/2}$ has the solution $y(t)^{3/2} = \frac{3}{2}t + \text{const}$. This suggests that $x_n^{3/2} \approx \frac{3}{2}n$.

2. From the previous observation, we ''linearize'' $x_n$ by considering $y_n := x_n^{3/2}$. This new sequence satisfies the following recurrence relation: $$ y_{n+1} = y_n \left( 1 + \frac{1}{y_n} \right)^{3/2} = y_n + \frac{3}{2} + \frac{3}{8y_n} + \mathcal{O}(y_n^{-2}). $$ As in @clark's answer, this immediately yields the asymptotics $y_n \sim \frac{3}{2}n$. Moreover, this type of recurrence relation is well-understood. See my blog posting, for instance.