6
$\begingroup$

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?

$\endgroup$
0
5
+50
$\begingroup$

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 $$u_n=nv_n^{-1/2}=\frac{3}{\sqrt{n}}\left(1+\frac65n^{-1}+o(n^{-1})\right)^{-1/2}=\frac{3}{\sqrt{n}}\left(1-\frac35n^{-1}+o(n^{-1})\right).$$

$\endgroup$
1
  • $\begingroup$ Thank you for rhis beatiful answer $\endgroup$ – Jane Jul 12 '20 at 9:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.