Lets1 = √(6),s2 = √(6+√(6)),s3 = √(6+√(6+√(6))), and in general definesn+1 = √(6+sn). Prove that (sn) converges, and find its limit.

*note: "√" means "the square root of"

Printable View

- Apr 8th 2009, 04:29 PMbearej50Monotone and Cauchy Sequences
Let

*s*1 = √(6),*s*2 = √(6+√(6)),*s*3 = √(6+√(6+√(6))), and in general define*s**n*+1 = √(6+*s**n*). Prove that (*s**n*) converges, and find its limit.

*note: "√" means "the square root of"

- Apr 8th 2009, 04:44 PMJhevon
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 $\displaystyle \lim s_{n + 1} = \lim s_n$, call this limit $\displaystyle L$.

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

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

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

you want the positive solution (why?) - Apr 8th 2009, 05:55 PMbearej50
I am still confused...

- Apr 8th 2009, 06:16 PMJhevon
- Apr 9th 2009, 06:11 AMbearej50
I know that I have to prove that s

*n*+1 > s*n*, which will prove that it is monotonically increasing. Then I would have to find an upper bound, U, for it and prove that s*n*< U and s*n*+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. - Apr 9th 2009, 07:40 AMPlato
Here is the base case: $\displaystyle s_1 = \sqrt 6 < \sqrt {6 + \sqrt 6 } = s_2 < 3$.

Say it is true for K:

$\displaystyle \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. - Apr 11th 2009, 03:16 PMxalk
O.K.

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

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

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

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

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

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

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

Up to now we have proved :

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