Let $u_{1} \in \,]0,\pi[$ and $u_{n+1}=(1+\frac 1n)\sin(u_n),\ \forall n >0$

1- Prove that $\displaystyle \lim_{n\to +\infty} u_n=0$

2- Prove that $\displaystyle u_{n}=\frac{3}{\sqrt{n}}-\frac{9}{5n\sqrt{n}}+o(\frac{1}{n\sqrt{n}})$

To prove 1.

It's easy to show that $u_n\in [0,\frac {\pi}2],\quad \forall n>1$

Put $l=\limsup u_n=\limsup u_{n+1}$, by result 1, and result 2 we have $$\displaystyle l=\limsup \big((1+\frac 1n) \sin(u_n)\big)=\limsup \sin(u_n) =\sin(\limsup u_n)=\sin(l)$$ But $\sin l=l$ where $l\in [0,\frac{\pi}2]$ implie $l=0$. Thus $\displaystyle \lim_{n\to +\infty} u_n=0$

any help to prove 2?


I will use the following consequence of the Stolz–Cesàro theorem: for any $a>0$, $$\lim_{n\to\infty}\frac{x_{n+1}-x_n}{n^{a-1}}=x\implies\lim_{n\to\infty}\frac{x_n}{n^a}=\frac{x}{a}.\label{stolz}\tag{1}$$

Now consider $v_n=(n/u_n)^2$. Then we have (and the reason for such a choice is seen from) $$v_{n+1}=v_n\left(\frac{u_n}{\sin u_n}\right)^2=v_n\left(1+\frac{u_n^2}{3}+o(u_n^2)\right)=v_n+\frac{n^2}{3}+o(n^2),\label{expan}\tag{2}$$ knowing that $u_n=o(1)$ as you do. Thus, $\displaystyle\lim_{n\to\infty}\frac{v_{n+1}-v_n}{n^2}=\frac13$ and, by \eqref{stolz}, $\displaystyle\lim_{n\to\infty}\frac{v_n}{n^3}=\frac19$.

This process can be continued. We just take more terms in \eqref{expan}: $$v_{n+1}=v_n\left(1+\frac{u_n^2}{3}+\frac{u_n^4}{15}+\frac{2u_n^6}{189}+\ldots\right)$$

So, let $v_n=(n^3/9)+w_n$ with $w_n=o(n^3)$. Then $$w_{n+1}-w_n=\frac{n^3-(n+1)^3}{9}+\frac{n^2}{3}+\frac{n^4}{15\left(\frac{n^3}{9}+w_n\right)}+o(n)=\frac{4n}{15}+o(n)\\\underset{\eqref{stolz}}{\implies}\lim_{n\to\infty}\frac{w_n}{n^2}=\frac{2}{15}\implies v_n=\frac{n^3}{9}+\frac{2n^2}{15}+o(n^2).$$

Further development is more subtle (try it). But the above is sufficient to show $$u_n=nv_n^{-1/2}=\frac{3}{\sqrt{n}}\left(1+\frac65n^{-1}+o(n^{-1})\right)^{-1/2}=\frac{3}{\sqrt{n}}\left(1-\frac35n^{-1}+o(n^{-1})\right).$$

  • $\begingroup$ Thank you for rhis beatiful answer $\endgroup$ – Jane Jul 12 '20 at 9:31

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