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Suppose $a>0$. Let $x_1=\sqrt{a}$, and define $x_{n+1}=\sqrt{a+x_n}$ for $n\ge 1$. I've already used induction to show that $x_n<1+\sqrt{a}$ for all $n$ and that $\{x_n\}$ is an increasing sequence. I know that it is bounded and must converge by the monotone convergence theorem. I'm just not sure how to find the limit.

Any suggestions?

TheMobiusLoops
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1 Answers1

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The trick is that if it converges, you can get at the limit $x = \sqrt{a + x}$. Now just solve for $x$.

Edit: In particular $x^2 - x - a = 0$, so that $x = \frac{1 \pm \sqrt{1 + 4a}}{2}$. Notice that $x > 0$, so we can discard the negative solution.

J. J.
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