Show that the sequence is monotone and bounded.

Question

Show that the sequence defined by $a_1=1$ and $a_n=\sqrt{3+a_{n-1}}$ for $n>1$ is monotone and bounded. Then find the limit of the sequence.

I'm supposed to do this using induction. I'm usually pretty good with induction proofs, but I do have that, assuming convergence, the limit will be

$L=\sqrt{3+L}$

And so

$L^2-L-3=0$

And blah blah quadratic equation, $\frac{\sqrt{13}+1}{2}$.

Special case of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ — YuiTo Cheng, Jul 11 '19 at 08:01

score 5 · Accepted Answer · answered May 06 '15 at 05:35

Monotonicity: We have $a_2 = 2 > 1 = a_1$. Now suppose that $a_{n} > a_{n-1}$. Then $a_{n+1} = \sqrt{3 + a_{n}} > \sqrt{3 + a_{n-1}} = a_n$. This shows that $a_n$ is monotonically increasing.

Boundedness: We have $a_1 < 3$. Suppose $a_n < 3$. Then $a_{n+1} = \sqrt{3 + a_{n}} < \sqrt{3 + 3} = \sqrt{6} < 3$. It follows that $a_n < 3$ for all $n$, so $a_n$ is bounded above.

Since a bounded, monotone sequence converges, your limit calculation is valid.

score 2 · Answer 2 · answered May 06 '15 at 05:34

The function $f(x) = \sqrt{x+3}$ is an increasing function on $(0,\infty)$, and since $a_2 = 2 > 1 = a_1 \Rightarrow a_n = f(a_{n-1}) > f(a_{n-2}) = a_{n-1}$, and it is monotonically increasing. You can show also that it is bounded above by $3$ by induction, thus is convergent and you have done the part of finding its limit $L$.

score 2 · Answer 3 · answered May 06 '15 at 05:44

First observe that $a_n \leqslant \frac{\sqrt{13} +1}{2}=\xi $ for all $n.$ Indeed, this is true for $n=1, $ and if we assume this inequality to be true for some $k$ then $$a_{k+1} =\sqrt{3+a_k}\leqslant \sqrt{3+\xi}=\sqrt{\xi^2} =\xi.$$ Next observe that $$u^2 -u-3\leqslant 0$$ for $u\in \left<\frac{1-\sqrt{13}}{2} ,\frac{1+\sqrt{13}}{2}\right>$ hence $$a_n^2 -a_n -3 \leqslant 0$$ and therefore $$a_n \leqslant \sqrt{a_n +3}= a_{n+1}$$ so the sequence $(a_n )$ is monotone.

score 2 · Answer 4 · answered May 06 '15 at 05:52

First prove by induction that $1\leq a_n\lt 3$. This is true for $a_1$ and assuming it is for $a_{n-1}$ one has $1\leq 2\leq a_n=\sqrt{a_n+3}\lt \sqrt{6}\lt 3$.

Let us now compute

$$a_{n+1}-a_n=\sqrt{a_n+3}-\sqrt{a_{n-1}+3}=\frac{a_n-a_{n-1}}{\sqrt{a_n+3}+\sqrt{a_{n-1}+3}}$$

The denominator is $2\leq D\leq 6$ and one has

$$\frac{a_n-a_{n-1}}{6}\leq a_{n+1}-a_n\leq \frac{a_n-a_{n-1}}{2}$$

We have $a_2-a_1=1\geq 0$ assuming $a_n-a_{n-1}\geq 0$ the above inequality shows it is the case for $a_{n+1}-a_n$.

We have proven that the sequence is monotone increasing and bounded.

Show that the sequence is monotone and bounded.

4 Answers4