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 $\displaystyle \lim s_{n + 1} = \lim s_n$, 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?)
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.
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.
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