The convergence of a recurrcively defined sequence.

Question

Let $a_1=\sqrt{2}$ and $a_n=\sqrt{2+a_{n-1}}$ determine the convergence of the sequence and find its limit.

I know the sequence converges to $2$ and i can show this informally. But I don't know how to prove that formally.

Answer 1

Show by induction that $a_n$ is increasing.

Then show by induction that $a_n$ is bounded above, e.g. try showing that $a_n \le 2$ for all $n$.

Then a theorem garantuees that $a_n$ has a (unique) limit $L$.

Then $L = \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} \sqrt{2 + a_n} = \sqrt{2+L}$, as the square root and addition are continuous.

Now solve $L = \sqrt{L+2}$. This will have 2 solutions of which one has to be eliminated.

Answer 2

Once proved the convergence by induction as suggested by Henno, you might be interested in seeing directly that the limit is $2$. So let's find $a_n$ (without the dependence from $a_{n-1}$) and then take the limit as $n\to \infty$. Here we have: $$ a_1=\sqrt{2}\\ a_2=\sqrt{2+\sqrt{2}}=\sqrt{\sqrt{2}(\sqrt{2}+1)}=\sqrt[4]{2}\sqrt{\sqrt{2}+1}\\ a_3=\sqrt{2+\sqrt[4]{2}\sqrt{\sqrt{2}+1}}=\sqrt{2^{1/4}\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}=2^{1/8}\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)} \\ a_4=\sqrt{2+2^{1/8}\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}}=2^{1/16}\sqrt{2^{7/8}+\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}} \\ a_5=\sqrt{2+2^{1/16}\sqrt{2^{7/8}+\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}}}= \\2^{1/32}\sqrt{2^{15/16}+\sqrt{2^{7/8}+\sqrt{\left(2^{3/4}+\sqrt{\sqrt{2}+1}\right)}}} \\ \vdots \\ a_n=2^{1/2^n}\sqrt{2^{1-1/2^{n-1}}+\sqrt{2^{1-1/2^{n-2}}+...+\sqrt{\sqrt{2}+1}}}.$$

The limit is $$\lim_{n\to \infty} a_n = \lim_{n\to \infty} 2^{1/2^n} \times \lim_{n\to \infty} \sqrt{2^{1-1/2^{n-1}}+\sqrt{2^{1-1/2^{n-2}}+...+\sqrt{\sqrt{2}+1}}}= \\ \sqrt{2+\sqrt{2+\sqrt{2+...}}}=L \\ 2+\sqrt{2+\sqrt{2+...}}=2+L=L^2 \\ L^2-L-2=0.$$ Finally, using the quadratic formula we find $L=-1 \vee L=2$. Since $L>0$, it is $L=2$.