Solve the limit problem of this sequence

calculuslimitsreal-analysis

Set the sequence \begin{aligned}x_{n+1}=\sin x_n,x_1=x>0,\end{aligned}
compute
\begin{aligned}\lim_{x\to\ 0^+}\frac{\ \sum_{n=1}^{\ \infty\ }\ x_n^3}x.\end{aligned}

For this recursive sequence, I can estimate $x_n\sim\sqrt{3/n}$
The specific process is as follows,

It is not difficult to $\lim\limits_{n\to \infty}x_n=0$($0＜x_1＜\frac\pi2$)

\begin{aligned}\frac1{{x^2}_{n+1}}-\frac1{{x^2}_n}&=\frac1{\sin ^2\left( x_n \right)}-\frac1{{x^2}_n}\\&\thicksim \frac1{\left( x_n-\frac{{x^3}_n}6 \right)^2}-\frac1{{x^2}_n}\\&=\frac1{{x^2}_n}\left( \frac1{\left(1-\frac{{x^2}_n}6 \right)^2}-1\right)\\&\thicksim \frac1{{x^2}_n}\left(\frac{{x_n}^2}3\right)\\&=\frac13\\\therefore \sum_{k=1}^n{\frac1{{x^2}_{k+1}}-\frac1{{x^2}_k}}&=\frac1{{x^2}_{n+1}}-\frac1{{x^2}_1}\\&=\frac{n}3\\\therefore x_n&\sim \sqrt{\frac3{n}}\,\,,n\to \infty \\\end{aligned}
I know from this that the series converges, but, the limit of summation I don't really have any good way to solve it.

Best Answer

Let $\sin^{\circ n}(x)$ denote the $n$-fold composition of $\sin$. Then we claim:

$$ \bbox[padding:5px;border:1px dotted navy;]{ \lim_{x \to 0^+} \frac{1}{x} \sum_{n=0}^{\infty} [\sin^{\circ n}(x)]^3 = 6. } \label{a}\tag{$*$} $$

The key lemma is as follows:

Lemma. For each $0 < A < 3 < B$, there exists $T > 0$ such that

$$ \sqrt{\frac{A}{t+n}} \leq \sin^{\circ n}\biggl( \sqrt{\frac{3}{t}} \biggr) \leq \sqrt{\frac{B}{t+n}} \label{1}\tag{1} $$

for all $ t \geq T$ and $n \geq 0$.

Let me first explain how this lemma leads to the desired conclusion \eqref{a}. Let $x > 0$, and let $t = 3/x^2$ so that $x = \sqrt{\frac{3}{t}}$ holds. Now fix $B > 3$, and let $T$ be as in the lemma. Then whenever $x \leq \sqrt{3/T}$, the lemma gives

\begin{align*} \frac{1}{x} \sum_{n=0}^{\infty} [\sin^{\circ n}(x)]^3 &\leq \frac{1}{x} \sum_{n=0}^{\infty} \biggl( \frac{B}{(3/x^2)+n}\biggr)^{3/2} \\ &\leq \sum_{n=0}^{\infty} \biggl( \frac{B}{3+nx^2}\biggr)^{3/2} x^2. \end{align*}

Taking $\limsup$ as $x \to 0^+$, this becomes

$$ \limsup_{x \to 0^+} \frac{1}{x} \sum_{n=0}^{\infty} [\sin^{\circ n}(x)]^3 \leq \int_{0}^{\infty} \biggl( \frac{B}{3+s}\biggr)^{3/2} \, \mathrm{d}s = \frac{2B^{3/2}}{\sqrt{3}}. $$

Since $B > 3$ is arbitrary and the $\limsup$ does not depend on $B$, letting $B \to 3^+$ gives

$$ \limsup_{x \to 0^+} \frac{1}{x} \sum_{n=0}^{\infty} [\sin^{\circ n}(x)]^3 \leq 6. $$

A similar argument shows that

$$ \liminf_{x \to 0^+} \frac{1}{x} \sum_{n=0}^{\infty} [\sin^{\circ n}(x)]^3 \geq 6, $$

proving the desired conclusion \eqref{a} given the lemma. $\square$

Proof of Lemma. Fix $0 < A < 3 < B$. Then by Taylor's theroem,

  • $ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120}\cos(\xi_1)$ for some $\xi_1 \in [0, x]$ and
  • $\frac{1}{(t+1)^{1/2}} = \frac{1}{t^{1/2}} - \frac{1}{2t^{3/2}} + \frac{3}{8 (t + \xi_2)^{5/2}} $ for some $\xi_2 \in [0, 1]$.

By absorbing the remainder term to the next smallest term in each of the Taylor formula, at the expense of nudging the coefficient a little bit, we can find $T > 0$ such that all the following conditions are satisfied:

  1. For all $0 \leq x \leq \sqrt{B/T}$, $$ x - \frac{x^3}{6} \leq \sin x \leq x - \frac{x^3}{2B} $$

  2. For all $t \geq T$, $$ \frac{1}{t^{1/2}} - \frac{1}{2t^{3/2}} \leq \frac{1}{(t+1)^{1/2}} \leq \frac{1}{t^{1/2}} - \frac{A}{6t^{3/2}} $$

  3. Both $x \mapsto x - \frac{x^3}{6}$ and $x \mapsto x - \frac{x^3}{2B}$ are increasing on the interval $0 \leq x \leq \sqrt{B/T}$,

  4. $0 \leq \sin x \leq x$ in the interval $0 \leq x \leq \sqrt{B/T}$.

Now we prove the assertion of the lemma by induction. Let $t \geq T$. Since \eqref{1} clearly holds in the base case $n = 0$, it suffices to establish the induction step.

Suppose \eqref{1} holds for $n \geq 0$. For simplicity, define $(x_n)$, $(a_n)$, $(b_n)$ by

$$ x_n = \sin^{\circ n}\biggl( \sqrt{\frac{3}{t}} \biggr), \qquad a_n = \sqrt{\frac{A}{t+n}}, \qquad b_n = \sqrt{\frac{B}{t+n}}. $$

Then \eqref{1} simplifies to $a_n \leq x_n \leq b_n$. Now using the properties above,

\begin{align*} x_{n+1} \leq x_n - \frac{x_n^3}{2B} &\leq b_n - \frac{b_n^3}{2B} \\ &= \sqrt{B} \left( \frac{1}{(t + n)^{1/2}} - \frac{1}{2(t + n)^{3/2}} \right) \\ &\leq \sqrt{\frac{B}{t+n+1}} = b_{n+1}. \end{align*}

Similarly,

\begin{align*} x_{n+1} \geq x_n - \frac{x_n^3}{6} &\geq a_n - \frac{a_n^3}{6} \\ &= \sqrt{A} \left( \frac{1}{(t + n)^{1/2}} - \frac{A}{6(t+n)^{3/2}} \right) \\ &\geq \sqrt{\frac{A}{t+n+1}} = a_{n+1}. \end{align*}

Therefore \eqref{1} holds for $n+1$ and the lemma follows. $\square$

Addendum. I did some computations, and it seems that we actually have

$$ \sin^{\circ n}\left(\sqrt{\frac{3}{t}}\right) = \sqrt{\frac{3}{t+n}}\left[ 1 + \mathcal{O}\left( \frac{\log(1 + n/t)}{t+n} \right) \right] $$

uniformly as $t, n \to \infty$. Since the computation is dirty and lengthy, I decided to use a weaker but easier-to-prove fact (the key lemma bove) instead.

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