On a nested sin sequence

Consider the sequence $x_n$ defined recursively as

$x_1=1 \quad, \quad x_{n+1}=\sin x_n$

Prove that $\lim \limits_{n \rightarrow +\infty} \sqrt{n} x_n = \sqrt{3}$ .

Solution

Lemma: If $a_n$ is a sequence for which $\displaystyle \lim_{n\to+\infty}(a_{n+1}-a_n)=a$ then

$\lim_{n\to + \infty}\frac{a_n}n=a.$

Proof: In Stolz theorem we set $x_{n}=a_{n+1}$ and $y_n=n$ .

It is easy to see that $x_n$ is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives

$\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{1}{3}$

Therefore

$\lim_{n\to+\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right)=\frac{1}{3}$

Now, due to the lemma we have $\lim\limits_{n\to+\infty} na_n^2 = 3$ and the result follows.

Remark : The asymptotic now follows to be $\displaystyle x_n \sim \sqrt{\frac{3}{n}}$ .

Problem: Find what inequality should $\beta$ satisfy such that the series

$\mathcal{S}=\sum_{n=1}^{\infty} x_n^\beta$

converges.

Read maker

One thought on “On a nested sin sequence”

Tolaso says:

June 16, 2019 at 11:59 am

We present a generalisation .

Claim: It holds that:

$x_n = \sqrt{\frac{3}{n}} - \frac{3^{3/2} \log n}{10 n^{3/2}} + \mathcal{O} \left ( n^{-3/2} \right )$

Proof: Left to the reader.

Log in to Reply