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I was working on the same question listed here and came up with the follwing proof with help of answers on that and other questions.

$\lim s_n$ converges so let $\lim s_n =s$.

Now $s_{n+1}= \sqrt{s_n+1}$ thus $\lim_{n \to \infty} s_{n+1} = \lim_{n \to \infty} \sqrt{s_n +1}$.

From here we get $s=\sqrt{s+1}$.

Simplifying gives: $s^2-s-1$. We do the quadratic formula and go without the extraneous solution and end up with $\lim_{n \to \infty} s_n = \frac{1+\sqrt{5}}{2}$

Now I get lost from the limits to plugging in s? I was using the answer here to help with my thought process. Why can we "substitute" $s$?

K Math
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That is because the square root function is continuous, so $$\lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}\bigl(\sqrt{s_{n}+1}\,\bigr)=\sqrt{\lim_{n\to\infty}(s_n+1)}=\sqrt{(\lim_{n\to\infty}s_n)+1}.$$

Bernard
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