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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. ${1+\sqrt{33}}\over{2}$

  2. ${1+\sqrt{32}}\over{2}$

  3. ${1+\sqrt{30}}\over{2}$

  4. ${1+\sqrt{29}}\over{2}$

How to proceed now. Thanks for any help.

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Trick: Let $X = \sqrt{ 7 + \sqrt{ 7 + ... } } $. We have $X = \sqrt{ 7 + X } $ and so $X^2 = 7 + X $. Now you solve the quadratic equation.

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    $\begingroup$ This is fine if you've proved it converges, but I'm not sure OP has done that. $\endgroup$ – Javier Aug 22 '15 at 17:18
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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 $a_{n+1} = \sqrt{7 + a_{n}} > \sqrt{7 + a_{n-1}} = a_{n}$.

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