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?

Medicim6

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} by induction also. For
$n=1,{a}_{1}=\sqrt{c}
If $c<\frac{1}{4}$, we're done since $LHS<0.
Otherwise, $c>\frac{1}{4}⇒2\sqrt{c}-1>0$, and squaring both sides: $4c-4\sqrt{c}+1<1+4c⇔-4\sqrt{c}<0$ is clearly true. Thus ${a}_{1}. Assume ${a}_{n}, you have ${a}_{n+1}=\sqrt{c+{a}_{n}}
$=\frac{1+2\sqrt{1+4c}+1+4c}{4}-c=\frac{1+\sqrt{1+4c}}{2}=M$
which is true by inductive step. So ${a}_{n}.
Thus the claim is verified. This follows that the limit exists and call it M. Then $M=\sqrt{c+M}⇒{M}^{2}=c+M⇒{M}^{2}-M-c=0⇒M=\frac{1±\sqrt{1+4c}}{2}$
Since ${a}_{n}>0,\mathrm{\forall }n\ge 1⇒M\ge 0⇒M=\frac{1+\sqrt{1+4c}}{2}$

Wendy Boykin

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
${y}_{±}=\frac{1±\sqrt{1+4c}}{2}$
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 .
$x
Thus, the sequence is bounded from above by y and is increasing. As y is the only possible limit, it converges to y.

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