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Let $a_n \in \Bbb{R}$ such that: $$ a_{n+1}=1+\frac{2}{a_n} \text{ and } a_1=1$$

Prove that $\lim_{n \to +\infty}a_n=2$

We have that $a_n \geq 1,\forall n \in \Bbb{N}$ thus $$\limsup_na_n \geq \liminf_na_n \geq 1$$

Let $\liminf_na_n=l$

Then we have that $l^2-l-2=0$ thus $l=2,-1$

But $a_n \geq 1$ thus $l=2$.

Applying the same argument we prove that $\limsup_na_n=2$

Also we can also derive a contradiction if we assume that the limit superior and inferior are infinite.

Is my attempt an efficient way to prove this statement?

Also can someone provide additional solutions?

Thank you in advance.

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  • $\begingroup$ define $b_n = \frac{a_n-2}{a_n+1}$ and show that it satisfies $b_{n+1} = -\frac12 b_n$. use this to deduce a closed form expression of $a_n$. $\endgroup$ Dec 28 '17 at 12:03
  • $\begingroup$ @achillehui...is my way correct..i will try your hint...Also is there any textbook for attacking and solving problems with recursive sequences? $\endgroup$ Dec 28 '17 at 12:05
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    $\begingroup$ Your argument won't work. Unlike limit, $\liminf a_n = \ell$ doesn't implies $\liminf(1 + \frac{2}{a_n}) = 1 + \frac{2}{\ell}$ directly (even when you assume $a_n > 0$). However for positive numbers $c_n$ bounded away from zero and infinity, we have $$\liminf \frac{1}{c_n} = \frac{1}{\limsup c_n}\quad\text{ and }\quad \limsup \frac{1}{c_n} = \frac{1}{\liminf c_n}$$ If $\limsup a_n = u$, then you can deduce from $a_{n+1} = 1 + \frac{2}{a_n}$ the consequence $$\ell = 1 + \frac{2}{u}\;\text{ and }\; u = 1 + \frac{2}{\ell} \implies (\ell-2)(\ell+1) = 0 \implies \ell = 2$$ $\endgroup$ Dec 28 '17 at 12:39
  • $\begingroup$ @achillehui...you are right..i had tottaly forgotten the two relations with limsup and liminf you mention..how can we overcome hte possibilty that the sequence is unbounded without finding an explicit formula? $\endgroup$ Dec 28 '17 at 12:48
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    $\begingroup$ The map $x \mapsto 1 + \frac{2}{x}$ send $(0,\infty)$ to $(1,\infty)$, the next iteration send it to $(1,3)$. If you start from any $a_1 \in (0,\infty)$, you have $a_n \in (1,3)$ for all $n > 2$. $\endgroup$ Dec 28 '17 at 12:52
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Banach's Fixed Point theorem also provides the limit. It states that the iteration $a_{n+1} = f(a_n)$ will converge to $a^* = f(a^*)$ if $|f'(a)| <1$ during the iteration.

Now $|f'(a)| = 2/a^2$ which in general will not be less than one. Taking two iteration steps however gives

$$ a_{n+2} = g(a_n) = 3 - \frac{4}{a_n+2} $$

and $|g'(a)| = \frac{4}{(a+2)^2}$ which is less than one for all $a_n >0$, which will always be the case over the iteration.

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Here we have where $a_{n+1}=f(a_n)$ where $$f(x)=1+\frac2x=\frac{x+2}{x}.$$ Maps of the form $x\mapsto (ax+b)/(cx+d)$ are fractional linear transformations and can be represented by matrices. If we let $A=\pmatrix{a&b\\c&d}$ and define $$A\bullet x=\frac{ax+b}{cx+d}$$ then $A\bullet (B\bullet x)=AB\bullet x$.

Here take $A=\pmatrix{1&2\\1&0}$. Then $a_{n+1}=A^n\bullet1$. We can diagonalise $A$: $$A=\pmatrix{1&2\\-1&1}\pmatrix{-1&0\\0&2}\pmatrix{1&2\\-1&1}^{-1}$$ and so $$A^{n-1}=\pmatrix{1&2\\-1&1}\pmatrix{(-1)^n&0\\0&2^n} \pmatrix{1&2\\-1&1}^{-1}.$$ Now it is straightforward to calculate Dr Graubner's formula.

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an explicit formula is given by $$a_n=-\frac{2+\left(\frac{-1}{2}\right)^n}{-1+\left(\frac{-1}{2}\right)^n}$$ so $$\lim_{n\to \infty}a_n=2$$

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Hint:

  • $a_n \ge 2 \implies a_{n+2} \ge 2, \ a_{n+2} \le a_n$

  • $a_n \le 2 \implies a_{n+2} \le 2, \ a_{n+2} \ge a_n$

  • $ 1 = a_1 \le a_3 \le a_5 \le \cdots \le 2 \le \cdots \le a_6 \le a_4 \le a_2 =3 $

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$a_{n+2}-a_n=\frac{4(a_n-a_{n-2})}{(a_n+2)(a_{n-2}+2)} - ①$

We can confirm the sign of $a_{n+2}-a_{n}$ and that of $a_{n}-a_{n-2}$ are same from ①.

For $a_1 < a_3$ and $a_4<a_2$, we can say $a_{2m-1} < a_{2m+1}$ and $a_{2m+2}<a_{2m}$ respectively.
Thus, $a_1<a_3<…<a_{2m-1}<…<2$, $a_2>a_4>…>a_{2m>}…>2$
$\{a_{2m-1}\}$ has upper bound and let $\alpha$ be its limit value. Also, $\{a_{2m}\}$ has lower bound and let $\beta$ be its limit value.

We can gain $\beta=1+\frac{2}{\alpha}-②$ and $\alpha=1+\frac{2}{\beta}-③$ from $a_{2m}=1+\frac{2}{a_{2m-1}}, a_{2m+1}=1+\frac{2}{a_{2m}}(m\rightarrow\infty)$

Finally, we can get $\alpha=\beta=2$ from ② and ③. This completes the proof.

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