Compute up to two decimal places the number$$ \sqrt{1+2 \sqrt{1+2 \sqrt{1+\cdots+2 \sqrt{1+2 \sqrt{1969}}}}} $$where the expression contains 1969 square roots.
Let$$ x_{n}=\sqrt{1+2 \sqrt{1+2 \sqrt{1+\cdots+2 \sqrt{1+2 \sqrt{1969}}}}} $$with the expression containing $n$ square root signs. Note that$$ x_{1}-(1+\sqrt{2})=\sqrt{1969}-(1+\sqrt{2})<50 . $$Also, since $\sqrt{1+2(1+\sqrt{2})}=1+\sqrt{2}$, we have $$ \begin{aligned} x_{n+1}-(1+\sqrt{2})=& \sqrt{1+2 x_{n}}-\sqrt{1+2(1+\sqrt{2})}=\frac{2\left(x_{n}-(1-\sqrt{2})\right)}{\sqrt{1+2 x_{n}}+\sqrt{1+2(1+\sqrt{2})}} \\ &<\frac{x_{n}-(1+\sqrt{2})}{1+\sqrt{2}} . \end{aligned} $$From here we deduce that$$ x_{1969}-(1+\sqrt{2})<\frac{50}{(1+\sqrt{2})^{1968}}<10^{-3}, $$and the approximation of $x_{1969}$ with two decimal places coincides with that of $1+\sqrt{2}=$ 2.41. This argument proves also that the limit of the sequence is $1+\sqrt{2}$.