Tuesday 15 May 2018

calculus - How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=frac{a_{n-1}+a_{n-2}}{2}$ converges to $frac23$?


How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ converges to $\frac23$?





If we analyse terms:
$$
0,1,\frac{1}{2},\frac{3}{4},\frac{5}{8},\cdots.
$$



I'm asked to do this using a previously proved theorem which says that




if you have two sequences ${b_n}$ and ${c_n}$ converging both to the same limit $L$, then the sequence $a_n$ defined as $b_1,c_1,b_2,c_2,b_3,\dots$ converge to $L$.





In this case, $b_n$ would be $0,\cfrac{1}{2},\cfrac{5}{8},\cfrac{21}{32},\dots$ and $c_n$ would be $1,\cfrac{3}{4},\cfrac{11}{16},\cfrac{43}{64},\dots$



From here it's seems all I need to do is prove that $b_n$ and $c_n$ converge to $\cfrac{2}{3}$ and then, by the theorem, $a_n$ converges to $\cfrac{2}{3}$.



I need to define them because I need to prove $b_n$ and $c_n$ are monotone and bounded, in order to use the monotone convergence theorem. I've got troubles when trying to define $b_n$ and $c_n$. Can anyone help me to define them?

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...