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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.

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