prove that $\{a_n\}_{n=1}^{\infty}$ is bounded and monotone increasing.

Question

I am trying to prove the sequence $\{a_n\}_{n=1}^{\infty}$, which is defined by $a_{n+1} = \frac {2(a_n+1)}{a_n+2}$, and $a_1=1$.

(1) prove that it is monotone increasing. ($a_{n+1} \ge a_n$)

proof by induction.

$P(1)$ : $a_2= \frac {2(1+1)}{2+1} = \frac 43 \ge a_1 = 1$

Suppose $P(k)$ holds true : $a_k \le a_{k+1}$.

$P(k+1)$ : $a_{k+2}= \frac {2(a_k+1)}{a_k+2}$. Then, how can I proceed from here??

(2) prove that it is bounded ($a^2_n<2$)

$P(1): a^2_1=1<2 $

Suppose $P(k)$ holds true: $a^2_k<2$

$P(k+1): a^2_{k+1}= (\frac {2(a_k+1)}{a_k+2})^2= \frac {2a_k^2+4a_k+2}{a^2_k+4a_k+4}$ I also don't know how to proceed from here.

Thank you in advance !!

Answer 1

Prove first it is bounded: as it is defined recursively by the function $$ f(x)=\frac{2(x+1)}{x+2}=2-\frac 2{x+2} $$ we see, since $ 0 < a_1 < 2$, that for all $n:\quad$ (i) $\;a_n>0$;$\quad$(ii) $\;a_n<2$.

Next, observe $f$ is a increasing function on $[0,2]$. Therefore $(a_n)$ is a monotonic sequence, and its monotonicity is determined by what happens between $a_1$ and $a_2$: $$a_2=\frac{2(1+1)}{1+2}=\frac43>a_1,$$ hence it is an increasing sequence.

Answer 2

Hints:

• $a_{n+1} = \dfrac {2(a_n+1)}{a_n+2}\leq\dfrac{2(a_n+2)}{(a_n+2)}=2~\forall~n\geq 1$

• $\dfrac{2(x+1)}{x+2}\geq x~\forall~x\in [0,\sqrt 2]$

The sequence converges to $\sqrt 2$

Answer 3

I'll play around and see what happens.

$a_{n+1} = \frac {2(a_n+1)}{a_n+2}, a_1=1 $

$\begin{array}\\ a_{n+1}^2-2 &= \dfrac {4(a_n+1)^2}{(a_n+2)^2}-2\\ &= \dfrac {4a_n^2+8a_n+4-2(a_n^2+4a_n+4)}{(a_n+2)^2}\\ &= \dfrac {4a_n^2+8a_n+4-2a_n^2-8a_n-8}{(a_n+2)^2}\\ &= \dfrac {2a_n^2-4}{(a_n+2)^2}\\ &= 2\dfrac {a_n^2-2}{(a_n+2)^2}\\ \end{array} $

So $a_{n+1}^2-2$ has the same sign as $a_n^2-2$. Since $a_1 = 1$, all $a_n$ satisfy $a_n^2 < 2$.

$\begin{array}\\ a_{n+1} -a_n &= \dfrac {2(a_n+1)}{a_n+2}-a_n\\ &= \dfrac {2a_n+2-a_n^2-2a_n}{a_n+2}\\ &= \dfrac {2-a_n^2}{a_n+2}\\ &> 0 \qquad\text{since } a_n^2 < 2\\ \end{array} $

Note that if $a_1^2 > 2$ then all $a_n^2 > 2$ and the $a_n$ are decreasing.

In either case, the $a_n$ are bounded and monotonic, so they approach a limit. If $L$ is this limit, since $a_{n+1} -a_n = \dfrac {2-a_n^2}{a_n+2} $, $\dfrac {2-a_n^2}{a_n+2} \to 0$ so $a_n^2 \to 2$.