How to prove the limit of composition of infinitely many sins

Question

I was trying to prove the limit for the sequence $$a_n=\sin \sin\cdots \sin 1$$ (the composition of n sins) It of course is monotonically decreasing and will converge at the point 0, but fail to prove it only by using the definition of limit. That for every epsilon we can find some n that $a_n$ is smaller than epsilon.

Do anyone know how to solve this or can somebody give me some clues

Answer 1

Let us first recall how the usual proof of $a_n \to 0$ goes:

$(a_n)$ is bounded and monotone decreasing, hence it has some limit $L \in [0, 1]$. This limit must solve the equation $L = \sin L$, which has the unique solution $L = 0$. Hence $a_n \to L$.

Note that the crucial ingredient is the completeness of $\mathbb{R}$, which guarantees that the limit of $(a_n)$ indeed exists without knowing the value of it. So it seems to me that OP is asking for a proof that does not (at least directly) require the completeness of $\mathbb{R}$. Now, to implement $\varepsilon$-$N$ argument, we need an explicit estimation of the decaying speed of $(a_n)$. The lemma below serves this purpose:

Lemma. Let $C = \frac{60}{19}$, and let $(a_n)$ be defined by $a_0 = 1$ and $a_{n+1} = \sin a_n$. Then for any $n \geq 0$, we have $$ 0 \leq a_n \leq \min\biggl\{ 1, \sqrt{\frac{C}{n+3}} \biggr\}. \tag{1} $$

Assuming this lemma, the $\varepsilon$-$N$ argument for $(a_n)$ is immediate:

Let $\varepsilon > 0$ be arbitrary, and let $N \geq 0$ be such that $\sqrt{\frac{C}{N+3}} < \varepsilon$ where $C$ is as in the lemma. Then by the above lemma, we have $$ |a_n - 0| = a_n \leq \sqrt{\frac{C}{n+3}} \leq \sqrt{\frac{C}{N+3}} < \varepsilon. $$ Therefore $a_n \to 0$.

So it remains to establish the lemma:

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Proof of Lemma. We first collect some useful observations:

1. For $0 \leq x \leq 1$, the power series $\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$ is an alternating series. So, any partial sum that is truncated just after a positive term must upper-bound the actual sum, yielding $$ \sin x \leq x - \frac{x^3}{3!} + \frac{x^5}{5!} \leq x - \frac{x^3}{3!} + \frac{x^3}{5!} = f(x), $$ where $f(x) = x - \frac{x^3}{2C}$ and $C = \frac{60}{19}$ as in the lemma.

2. For $ t > 0$, Taylor's theorem applied to $s \mapsto \frac{1}{\sqrt{t+s}}$ tells that there exists $\xi \in [0, 1]$ satisfying $$ \frac{1}{\sqrt{t+1}} = \frac{1}{\sqrt{t}} - \frac{1}{2t^{3/2}} + \frac{3}{8(t+\xi)^{5/2}} \geq \frac{1}{\sqrt{t+1}} = \frac{1}{\sqrt{t}} - \frac{1}{2t^{3/2}}. $$

3. By investigating the derivative, it is easy to check that the function $f(x) = x - \frac{x^3}{2C}$ defined above is increasing in $[0, \sqrt{C/3}]$.

Now we are ready to prove the lemma by mathematical induction:

• Base Step. When $n = 0$, $\text{(1)}$ clearly holds.

• Inductive Step. Suppose $\text{(1)}$ holds for $n \geq 0$. Since $0 \leq \sin x \leq x$ for $x \in [0, 1]$, we immediately have $0 \leq a_{n+1} \leq a_n \leq 1$. Moreover, using the monotonicity of $f$ and the inductive hypothesis $a_n \leq \sqrt{\frac{C}{n+3}}$, we get

\begin{align*} a_{n+1} \leq f(a_n) \leq f\biggl( \sqrt{\frac{C}{n+3}} \biggr) &= \sqrt{C} \left( \frac{1}{\sqrt{n+3}} - \frac{1}{2(n+3)^{3/2}} \right) \\ &\leq \sqrt{\frac{C}{(n+1)+3}}. \end{align*}

Therefore the induction hypothesis is satisfied and we are done. $\square$