real analysis - Show that series $a_{n+1} = sqrt{1+sqrt{a_n}}$ converges

Let $a_1$ = $\sqrt{2}$ and $a_{n+1} = \sqrt{1+\sqrt{a_n}}$ for all natural numbers n.
Show that the sequence $a_n$ converges

I've been doing problems which use the Root/Ratio test. I don't see how I can use that here.
I started with $\sqrt2$ < $2$, and $\sqrt{a_1}$ < $1$ and tried to reason out the series. That doesn't work. How would I show that it converges to some value $L$?

Answer

The sequence is monotone increasing. Show that it is bounded from above and you are done.

Monotonicity. Proof by induction. We have $$a_1 - a_0 = \sqrt{1 + \sqrt{2}} - \sqrt{2} \approx 0.134 > 0$$ Hence assume the sequence is monotone increasing for $n > 0$. We have $$a_{n + 1} - a_n = \sqrt{1 + \sqrt{a_n}} - \sqrt{1 + \sqrt{a_{n-1}}} = \frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{1 + \sqrt{a_n}} + \sqrt{1 + \sqrt{a_{n-1}}}} = \frac{a_n - a_{n-1}}{\left( \sqrt{1 + \sqrt{a_n}} + \sqrt{1 + \sqrt{a_{n-1}}}\right)\left( \sqrt{a_n} + \sqrt{a_{n-1}}\right)} > 0$$ since by the induction hypothesis $a_n - a_{n-1} > 0$ and the sequence is for sure positive (one can show by induction, that the sequence is bounded below by $1$).

Boundedness. The sequence is bounded above by $2$ (this is not sharp, but its an upper bound). Obviously, $\sqrt{2} \leqslant 2$. Assume, $a_n \leqslant 2$. Then $$a_{n + 1} = \sqrt{1 + \sqrt{a_n}} \leqslant \sqrt{1 + \sqrt{2}} \leqslant \sqrt{4} = 2$$ since the root function is monotone increasing and of course $\sqrt{2} \leqslant 3$.

Hence we can conclude, that the sequence converges.

Limit. If you want to calculate $\lim_{n \to \infty} a_n$ we stick to the following trick. Denote the limit by $L$. Then it must hold $$L = \lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{1 + \sqrt{a_n}} = \sqrt{1 + \sqrt{L}}$$ since the root function is continuous. Hence we have to solve $$L = \sqrt{1 + \sqrt{L}}$$ or equivalently $$L^4 - 2L^2 - L + 1 = 0$$ However this is not so easy to solve. Perhaps, someone has a good idea to solve this, but using Wolfram, we get the two roots $$L_1 \approx 0.524829 \qquad L_2 \approx 1.4902$$ So for sure $$\lim_{n \to \infty} a_n \neq L_1$$ since the sequence is monotone increasing and bounded below by $1$. Thus $$\lim_{n \to \infty} a_n = L_2 \approx 1.4902 $$