Show that the sequence ${a_n}$ converges where $a_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}}$ for $n\geq 1$. The original question was to determine whether the sequence converges, but I have checked for extremely high values of $n$ and it seems as though it does converge. This lead me to wonder if there was an "easy" method of showing that the sequence is bounded (since it is monotone, it then follows that it converges). Thanks.

Term by term, is less than
$$\sqrt{1+\sqrt{2^1+\sqrt{2^2+\sqrt{2^3...}}}}=\sqrt{1+\sqrt{2}\sqrt{1+2^{-1}\sqrt{2^2+\sqrt{2^3...}}}}$$
$$<\sqrt{1+\sqrt{2}\sqrt{1+2^{-0.5}\sqrt{2^2+\sqrt{2^3+...}}}}$$
$$=\sqrt{1+\sqrt{2}\sqrt{1+\sqrt{2+2^{-1}\sqrt{2^3...}}}}$$
$$\cdots$$
$$<\sqrt{1+\sqrt{2}\sqrt{1+\sqrt{2+\sqrt{2^2...}}}}$$
So we now have to show that $a_0=1$,$a_{n+1}=\sqrt{1+\sqrt{2}a_{n}}$ converges(this series is strictly greater than the last equation).We proceed by induction.
Assume $a_n\le\frac{1+\sqrt3}{\sqrt2}$, then
$$a_{n+1}\le\sqrt{2+\sqrt{3}}=\frac{1+\sqrt3}{\sqrt2}$$
And since our sequence is monotone increasing and $1\le\frac{1+\sqrt3}{\sqrt2}$ the result follows. 
Note: The number $1.757933$ of ncmathsadist is fairly close to this limit, $\approx 1.931851$
