Find $\lim_{n \to \infty}\frac{x_n}{\sqrt{n}}$ where $x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}$

Question

Assume a positive sequence $\{x_n\} $ satisfies $$x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}.$$ Find $\lim\limits_{n \to \infty}\dfrac{x_n}{\sqrt{n}}$.

Assume the limit we want is $L$. Then by Stolz theorem, one can obtain \begin{align*} L&=\lim_{n \to \infty}\frac{x_n}{\sqrt{n}}=\lim_{n \to \infty}\frac{x_{n+1}-x_n}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty}\frac{2n\sqrt{n}}{x_1+x_2+\cdots+x_n}\\ &=\lim_{n \to \infty}\frac{2n\sqrt{n}-2(n-1)\sqrt{n-1}}{x_n}=\lim_{n \to \infty}\frac{3\sqrt{n}}{x_n}=\frac{3}{L}, \end{align*} which implies $L=\sqrt{3}$. But how to prove the limit exists?

Answer 1

This is a community-wiki answer illustrating Iosif Pinelis's solution (with some simplifications).

• If you like this, please give kudos to his original solution as well.
• Also, feel free to improve this answer as you please!

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Setting. We will write $s_n = \sum_{k=1}^{n} x_k$, so that the recurrence relation takes the form

$$ x_{n+1} = x_n + \frac{n}{s_n}. \tag{RE} $$

Also, let $\alpha$ and $\beta$ by

$$ \alpha = \liminf_{n\to\infty} \frac{x_n}{\sqrt{n}} \qquad\text{and}\qquad \beta = \limsup_{n\to\infty} \frac{x_n}{\sqrt{n}}. $$

Our goal is to prove that $\alpha = \beta = \sqrt{3}$. To make use of these quantities, we will frequently utilize the following inequalities:

Lemma. Let $(a_n)$ be any sequence of real numbers, and let $p > 0$. Then $$ \color{navy}{\liminf_{n\to\infty} \frac{a_n}{n^p}} \geq \liminf_{n\to\infty} \frac{a_n - a_{n-1}}{pn^{p-1}} \tag{SC1} $$ and $$ \color{navy}{\limsup_{n\to\infty} \frac{a_n}{n^p}} \leq \limsup_{n\to\infty} \frac{a_n - a_{n-1}}{pn^{p-1}}. \tag{SC2} $$

Proof. This is an immediate consequence of the Stolz–Cesàro theorem together with the asymptotic formula $n^p - (n-1)^p \sim pn^{p-1}$.

Step 1. Since $(x_n)$ is increasing, we know that $s_n \leq n x_n$. Then by the lemma,

\begin{align*} \alpha^2 = \liminf_{n\to\infty} \frac{x_n^2}{n} &\geq \liminf_{n\to\infty} \left( x_{n+1}^2 - x_n^2 \right) \tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\ &\geq \liminf_{n\to\infty} \frac{2nx_n}{s_n} \tag*{by $\color{#2E8B57}{\text{(RE)}}$} \\ &\geq 2. \tag*{$\because \ s_n \leq n x_n$} \end{align*}

Now, by the Stolz–Cesàro theorem again,

\begin{align*} \beta = \limsup_{n\to\infty} \frac{x_n}{\sqrt{n}} &\leq \limsup_{n\to\infty} \frac{x_{n+1} - x_n}{\frac{1}{2}n^{-1/2}} \tag*{by $\color{#2E8B57}{\text{(SC2)}}$} \\ &= \limsup_{n\to\infty} \frac{2n^{3/2}}{s_n} \tag*{by $\color{#2E8B57}{\text{(RE)}}$} \\ &= \left[ \liminf_{n\to\infty} \frac{s_n}{2n^{3/2}} \right]^{-1} \tag{1} \\ &\leq \left[ \liminf_{n\to\infty} \frac{x_n}{3\sqrt{n}} \right]^{-1} = \frac{3}{\alpha}. \tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \end{align*}

These altogether show that $0 < \alpha \leq \beta < \infty$.

(Remark. Starting from $\alpha$ and applying a similar argument as above, we can show that $\alpha \beta = 3$. However, this does not determine the value of $\alpha$ and $\beta$. So, we will not bother to prove this.)

Step 2. Using the recurrence relation $\color{#2E8B57}{\text{(RE)}}$, we find that

\begin{align*} s_{n+1}^2 - 2s_n^2 + s_{n-1}^2 &= (s_n + x_{n+1})^2 - 2s_n^2 + (s_n - x_n)^2 \\ &= 2 s_n(x_{n+1} - x_n) + x_{n+1}^2 + x_n^2 \\ &= 2n + x_{n+1}^2 + x_n^2. \tag{2} \end{align*}

Using this, we get

\begin{align*} \liminf_{n\to\infty} \frac{s_n^2}{n^3} &\geq \liminf_{n\to\infty} \frac{s_{n+1}^2 - s_n^2}{3n^2} \tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\ &\geq \liminf_{n\to\infty} \frac{s_{n+1}^2 - 2s_n^2 + s_{n+1}^2}{6n} \tag*{by $\color{#2E8B57}{\text{(SC1)}}$} \\ &\geq \frac{1}{3} + \frac{1}{6} \biggl( \liminf_{n\to\infty} \frac{x_{n+1}^2}{n} \biggr) + \frac{1}{6} \biggl( \liminf_{n\to\infty} \frac{x_n^2}{n} \biggr) \tag*{by $\color{#2E8B57}{\text{(2)}}$} \\ &= \frac{1 + \alpha^2}{3}. \end{align*}

Plugging this into $\color{#2E8B57}{\text{(1)}}$,

\begin{align*} \beta \leq \left[ \liminf_{n\to\infty} \frac{s_n}{2n^{3/2}} \right]^{-1} \leq 2\sqrt{\frac{3}{1+\alpha^2}}. \tag{3} \end{align*}

A similar calculation also shows that

$$ \alpha \geq 2\sqrt{\frac{3}{1+\beta^2}}. \tag{4} $$

Step 3. Define $f(x) = 2\sqrt{\frac{3}{1+x^2}}$, and note that $f$ is decreasing on $[0, \infty)$. So, using $\color{#2E8B57}{\text{(3)}}$ and $\color{#2E8B57}{\text{(4)}}$, we get

$$ \beta \leq f(\alpha) \implies f(f(\alpha)) \leq f(\beta) \qquad\text{and}\qquad \alpha \geq f(\beta) \implies f(\alpha) \leq f(f(\beta)) $$

and hence

$$ \beta \leq f(f(\beta)) \qquad\text{and}\qquad f(f(\alpha)) \leq \alpha. $$

It is not hard to check that these inequalities yield $\beta \leq \sqrt{3} \leq \alpha$:

So, using the obvious relation $\alpha \leq \beta$, we conclude that

$$ \sqrt{3} \leq \alpha \leq \beta \leq \sqrt{3} $$

and therefore $\alpha = \beta = \sqrt{3}$.