# Thread: Monotone and Cauchy Sequences

1. ## Monotone and Cauchy Sequences

Let s1 = (6), s2 = (6+(6)), s3 = (6+(6+(6))), and in general define sn+1 = (6+sn). Prove that (sn) converges, and find its limit.

*note: "√" means "the square root of"

2. Originally Posted by bearej50
Let s1 = (6), s2 = (6+(6)), s3 = (6+(6+(6))), and in general define sn+1 = (6+sn). Prove that (sn) converges, and find its limit.

*note: "√" means "the square root of"

show that the sequence is monotonic (increasing) and bounded. this will show that it has a limit (by what theorem?) you can prove both pretty easily by induction. do you see what to prove for each of these?

to find the limit, note that $\lim s_{n + 1} = \lim s_n$, call this limit $L$.

then, since $s_{n + 1} = \sqrt{6 + s_n}$ we have

$\lim s_{n + 1} = \lim \sqrt{6 + s_n}$

$\Rightarrow L = \sqrt{6 + L}$

you want the positive solution (why?)

3. I am still confused...

4. Originally Posted by bearej50
I am still confused...
it would help if you say where specifically you are having trouble.

take one part at a time. lets show that it is increasing. what have you done to this end?

5. I know that I have to prove that sn+1 > sn, which will prove that it is monotonically increasing. Then I would have to find an upper bound, U, for it and prove that sn < U and sn+1 < U. This would prove that it is bounded. Then, by the monotone convergence theorem, the sequence must be convergent and have a limit.
I am having trouble putting this all together in a proper proof.

6. Here is the base case: $s_1 = \sqrt 6 < \sqrt {6 + \sqrt 6 } = s_2 < 3$.

Say it is true for K:
$\begin{gathered}
s_K < s_{K + 1} < 3 \hfill \\
6 + s_K < 6 + s_{K + 1} < 9 \hfill \\
\underbrace {\sqrt {6 + s_K } }_{s_{K + 1} } < \underbrace {\sqrt {6 + s_{K + 1} } }_{s_{K + 2} } < 3 \hfill \\
\end{gathered}$

Done by induction: increasing and bounded.

7. Originally Posted by bearej50
I know that I have to prove that sn+1 > sn, which will prove that it is monotonically increasing. Then I would have to find an upper bound, U, for it and prove that sn < U and sn+1 < U. This would prove that it is bounded. Then, by the monotone convergence theorem, the sequence must be convergent and have a limit.
I am having trouble putting this all together in a proper proof.
O.K.

Lets take the difference $S_{n+1}-S_{n} =\sqrt{6+s_{n}}-s_{n}$.................................................. ......................................1

Now if you multiply (1) by $\frac{\sqrt{6+s_{n}}+s_{n}}{\sqrt{6+s_{n}}+s_{n}}$,and do couple of calculations you get the formula:

$S_{n+1}-S_{n} =-\frac{(s_{n}+2)(s_{n}-3)}{\sqrt{6+s_{n}}+s_{n}}$.................................................. .....................................2

In this formula all the terms are +ve except the factor $s_{n}-3$.

But since we can prove that $s_{n}<3$ for all n, then by using (2) we get $S_{n+1}-S_{n}>0$.

Hence the sequence is increasing and bounded from above by 3.

So it is left to you to prove $s_{n}<3$ for all ,n and $s_{n}>0$ for all ,n.

Up to now we have proved :

a) IF the sequence converges ,then it must converge to 3 (because $s_{n}>0$ for all ,n)

b) The sequence converges.

BY (a) and (b) and using M.Ponens ,we have:

The limit of the sequence is 3.

Note usually the finding of a limit for this type of sequences consists of those two parts.

In the first part WE Assume that the sequence converges,and find its limit .
In the 2nd part we prove convergence