TITLE: Where does this sequence $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.... converge?
QUESTION [5 upvotes]: The  given  sequence  is $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.....and  so  on. 
the  sequence  is  increasing  so  to  converge  must  be  bounded  above.Now  looks  like  they  would not  exceed  7. The  given  options  are 

${1+\sqrt{33}}\over{2}$
${1+\sqrt{32}}\over{2}$
${1+\sqrt{30}}\over{2}$
${1+\sqrt{29}}\over{2}$

How  to  proceed now. 
Thanks  for  any  help.

REPLY [4 votes]: Here's a quick proof that the sequence converges.
The sequence is given by the recursive formula $a_0 = 0$, and $a_{n+1} = \sqrt{7 + a_n}$.
Using the method suggested by the previous poster, let $L$ be the positive solution to the equation $x^2 = x + 7$. 
We can prove that $a_n < L$ for all $L$ by induction. It is clear that $a_0 = 0 < L$. Then if $a_n < L$ we have $a_{n+1} = \sqrt{7 + a_n} < \sqrt{7 + L} = L$. 
We can also prove that the sequence is increasing. Since $a_1 = \sqrt{7}$, we have $a_1 > a_0$. Now, if $a_n > a_{n-1}$, we have $a_{n+1} = \sqrt{7 + a_{n}} > \sqrt{7 + a_{n-1}} = a_{n}$.