Cfrgtkky

The question I'm currently struggling with is:

Consider the following recursively defined sequence:

$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$

a) Show by induction that $0le a_n le 1 $ for all n

b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.

c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.

d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$

I've managed to complete part a) where I did:

Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$

Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$

Inductive Step - Here I tried to prove it by making $n=k+1$ which meant

$a_{k+2}= frac{a_k+1^2+4}{5}$

which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$

and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$

Part b) seems relatively simple as you can prove

$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:

$a_{k+1}= frac{a^2_k+4}{5} $

Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.

I did the basis and my inductive hypothesis and at the inductive step I got to:

$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this

Would anyone mind helping me with part c) and d)?

It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$

which is true by a).

The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.

The convergence (d) then follows from the monotone convergence theorem.

b) The argument is that at the fixed point, which has the value A,
$$a_{k+1}=a_{k}=A$$

So
$$a_{k+1}=frac{a_k^2+4}{5}$$
becomes
$$A=frac{A^2+4}{5}$$
$$5A=A^2+4$$
$$A^2+5A+4=0$$

a) By the assumption of the induction we obtain:
$$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
which says that $a$ increases and bounded.

Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.

Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.

I like the following way:
$$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
Which says $$lim_{nrightarrow+infty}a_n=1.$$