Assuming convergence of the following series, find the value of $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$

I was advised to proceed with this problem through substitution but that does not seem to help unless I am substituting the wrong parts. If i substitute the $6$, well then i am just stuck with above.

Any ideas on how to proceed. Also, what is the purpose of stating that it is convergent.


marked as duplicate by MJD, Claude Leibovici, Jyrki Lahtonen Oct 7 '15 at 6:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ @GAVD I think this question quite is different from that one since it involves no unknown, just a simple "evaluate this expression...". $\endgroup$ – BigbearZzz Oct 7 '15 at 3:42

Let $x=\sqrt{6+\sqrt{6+\sqrt{6+...}}}$, then observe that $x=\sqrt{6+x}$. Squaring both sides yields $$ x^2=x+6 $$ , which is a quadratic formula. Solve it normally and choose the wise answer out of the 2 roots.

  • $\begingroup$ I see, so I would get -2 and 3. I would have to eliminate -2 since we cant have a negative number within the square root. $\endgroup$ – Caddy Heron Oct 7 '15 at 3:43
  • $\begingroup$ Exactly. I have a question that you might be interested in: What if BOTH roots are positive? Which one will you choose? :) $\endgroup$ – BigbearZzz Oct 7 '15 at 3:51
  • $\begingroup$ Interesting, I would say pick none but since its convergent, i would have to pick one? So i'm not sure. $\endgroup$ – Caddy Heron Oct 7 '15 at 3:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.