Given a recursion $a_{n+ 1}= \dfrac{1}{2}\left ( a_{n}+ \dfrac{1}{a_{n}} \right )$ with $a_{1}> 0.$ Find the $\mathcal{o}\left ( n \right )$ so that
$$a_{n}\sim\frac{\mathcal{o}\left ( n \right )}{n}+ \mathcal{O}\left ( \frac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$
Remark. The original one is proving $\lim a_{n}= 1.$ I see two approaches but which are kind of luckily nonsense and unuseful here.
-
We have that
$$a_{n+ 1}- 1= \frac{1}{2}\left ( a_{n}+ \frac{1}{a_{n}} \right )- 1= \frac{1}{2}\left ( a_{n}- 1 \right )^{2}+ \mathcal{O}\left ( \left ( a_{n}- 1 \right )^{2} \right )\,{\rm as}\,a_{n}\rightarrow 1$$
We easily find the $\mathcal{o}\left ( n \right )$ but actually I find $\mathcal{o}\left ( n \right )$ to prove that $\lim a_{n}= 1,$ that's a bad idea ! -
From the example * Estimations of some new recurrence sequences *, for the recursion $a_{1}= 1$ and $a_{n+ 1}= \sqrt{a_{n}^{2}+ a_{n}}:$
$$a_{n}\sim\frac{1}{2}\left ( n+ \frac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$
I need to the help to change the problem of the disconvergences to the convergence problem by using $n$ cleverly, I know it's too inconvenient, but I can't think such a better way at all. Thanks for all your nice and valuable comments a real lot !
Best Answer
We have: $$a_{n+1}-1 = \frac{1}{2}\frac{(a_n-1)^2}{a_n}$$ $$a_{n+1}+1 = \frac{1}{2}\frac{(a_n+1)^2}{a_n}$$ Hence $$\frac{a_{n+1}-1}{a_{n+1}+1}=\left( \frac{a_{n}-1}{a_{n}+1} \right)^2$$ We deduce that
$$\frac{a_{n}-1}{a_{n}+1} = \left( \frac{a_{1}-1}{a_{1}+1} \right)^{2^{n-1}}$$ or the closed form expression of $a_n$ $$a_n =\frac{1-\left( \frac{a_{1}-1}{a_{1}+1} \right)^{2^{n-1}}}{1+\left( \frac{a_{1}-1}{a_{1}+1} \right)^{2^{n-1}}}$$