oliviayychengwh

2022-01-15

Investigate the convergence or divergence of the sequence

${a}_{1}=\sqrt{c},{a}_{2}=\sqrt{c+{a}_{1}},{a}_{n+1}=\sqrt{c+{a}_{n}},c>0$ .

I proved it is increasing. But I could not prove it is bounded above or unbounded. I guess it is divergence sequence. How can I do this?

I proved it is increasing. But I could not prove it is bounded above or unbounded. I guess it is divergence sequence. How can I do this?

Medicim6

Beginner2022-01-16Added 33 answers

Claim: $a}_{n$ is increasing and bounded above by

$M=\frac{1+\sqrt{1}+4c}{2}$ . Also $\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=M$

Proof: By induction on$n\ge 1$ .

$a}_{2}=\sqrt{c+\sqrt{c}}>\sqrt{c}={a}_{1$ . Assume $a}_{n}>{a}_{n-1$ , we show $a}_{n+1}>{a}_{n$ .

But$a}_{n+1}=\sqrt{c+{a}_{n}}>\sqrt{c+{a}_{n-1}}={a}_{n$ by the inductive step. So by induction ${a}_{n+1}>{a}_{n},\mathrm{\forall}n\ge 1$ . We show that ${a}_{n}<M,\mathrm{\forall}n\ge 1$ by induction also. For

$n=1,{a}_{1}=\sqrt{c}<M\iff \sqrt{c}<\frac{1+\sqrt{1+4c}}{2}\iff 2\sqrt{c}-1<\sqrt{1+4c}$

If$c<\frac{1}{4}$ , we're done since $LHS<0<RHS$ .

Otherwise,$c>\frac{1}{4}\Rightarrow 2\sqrt{c}-1>0$ , and squaring both sides: $4c-4\sqrt{c}+1<1+4c\iff -4\sqrt{c}<0$ is clearly true. Thus ${a}_{1}<M$ . Assume ${a}_{n}<M$ , you have ${a}_{n+1}=\sqrt{c+{a}_{n}}<M\iff {a}_{n}<{M}^{2}-c$

$=\frac{1+2\sqrt{1+4c}+1+4c}{4}-c=\frac{1+\sqrt{1+4c}}{2}=M$

which is true by inductive step. So${a}_{n}<M,\mathrm{\forall}n\ge 1$ .

Thus the claim is verified. This follows that the limit exists and call it M. Then$M=\sqrt{c+M}\Rightarrow {M}^{2}=c+M\Rightarrow {M}^{2}-M-c=0\Rightarrow M=\frac{1\pm \sqrt{1+4c}}{2}$

Since$a}_{n}>0,\mathrm{\forall}n\ge 1\Rightarrow M\ge 0\Rightarrow M=\frac{1+\sqrt{1+4c}}{2$

Proof: By induction on

But

If

Otherwise,

which is true by inductive step. So

Thus the claim is verified. This follows that the limit exists and call it M. Then

Since

Wendy Boykin

Beginner2022-01-17Added 35 answers

First, you need to study the fixed points of the function

$f\left(x\right)=\sqrt{c+x}$

as only these can be the limit of the sequence

$f\left(0\right),f\left(f\left(0\right)\right),f\left(f\left(f\left(0\right)\right)\right),\dots$

Solving$y=\sqrt{c+y}$ , we find

$y}_{\pm}=\frac{1\pm \sqrt{1+4c}}{2$

The sequence starts at$x=0$ and is nonnegative, so the only possible candidate for the limit is

$y=\frac{1+\sqrt{1+4c}}{2}$

Now, we need to prove that if$0<x<y\text{}\text{then}\text{}xf\left(x\right)y$ .

$x<f\left(x\right)=\sqrt{c+x}<\sqrt{c+y}=f\left(y\right)=y$

Thus, the sequence is bounded from above by y and is increasing. As y is the only possible limit, it converges to y.

as only these can be the limit of the sequence

Solving

The sequence starts at

Now, we need to prove that if

Thus, the sequence is bounded from above by y and is increasing. As y is the only possible limit, it converges to y.