Vtyjkyuk

So I got this question and I was able to do up to (III) (a) however part b has been pretty difficult for me to complete and I can't seem to get it right . Any help ?

I tried considering $a_n - a_n+1$ and I got the required proof but I don't feel as though this method is logical enough

$a_n - a_n+1 < 0$

$a_n < a_n+1$

Item III.a. we use II. result $a_n+1 - 2 = 2 - frac124 + a_n$, and $a_n < 2 iff a_n + 4 < 6$, then $frac124 + a_n > 2$ Therefore $a_n+1 - 2 = 2 - frac124 + a_n < 0 iff a_n+1 < 2$.

Guide:

• Prove that $a_n>0$ if $a_1>0$.


• Compute $a_n+1-a_n$ in terms of $a_n$, your goal is to check that the expression is positive.


• check that the denominator is positive.


• Show that the numerator is positive using the property that $0<a_n<2$.

Alt. hint: $;a_n+1=dfrac4(a_n+1colorred+3-3)a_n+4=4 - dfrac12a_n+4,$, and therefore:

$$requirecancel
a_n+1-a_n=left(cancel4 - dfrac12a_n+4right) - left(cancel4 - dfrac12a_n-1+4right) = dfrac12(a_n-a_n-1)(a_n+4)(a_n-1+4)
$$

It follows that $,a_n+1-a_n,$ and $,a_n-a_n-1,$ have the same sign, so the entire sequence is monotonic, and the direction of monotonicity is given by the sign of $,a_2-a_1,$, in this case positive so the sequence is increasing.

So you can check your solutions.

We get $a_2=frac2011$ and $a_3=frac3116$.

For II):

It is $a_n+1-2=frac4(1+a_n)4+a_n-2=frac4+4a_n-8-2a_n4+a_n=frac2a_n-44+a_n=frac2(a_n-2)4+a_n$

For III):

It is $a_n+1=frac4(1+a_n)4+a_nstackrela_n<2<frac4(1+2)4+a_n=frac124+a_nstackrela_n<2<frac124+2=2$. Hence $a_n+1<2$.

Finally:

We show, that $a_n+1-a_n>0$

We have $frac4(1+a_n)1+a_n-a_n=frac4(1+a_n)-a_n(1+a_n)1+a_n=frac4-a_n^24+a_n=frac(2+a_n)(2-a_n)1+a_n>0$

Because $2+a_n>0$ and $1+a_n>0$ since $a_n>0$ (sequence of positiv numbers)
Also $2-a_n>0Leftrightarrow 2>a_n$ which is true.

Every factor greater than zero means the product is greater than zero.