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.

By repeatedly rationalizing the numerators and using that $a_n\geq1$ for all $n$, $$ a_{n+1}-a_n=\frac{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}-\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}{a_{n+1}+a_n}\\ \leq\frac{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}-\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}{2}\\ =\frac12\,\frac{\sqrt{3+\sqrt{4+\cdots\sqrt{n+1}}}-\sqrt{3+\sqrt{4+\cdots\sqrt{n}}}}{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}+\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}\\ \leq\frac12\,\frac{\sqrt{3+\sqrt{4+\cdots\sqrt{n+1}}}-\sqrt{3+\sqrt{4+\cdots\sqrt{n}}}}{2\sqrt2}\\ \leq\cdots\\ \leq\frac{\sqrt{n+1}}{2^{n+1}}\leq\frac{n+1}{2^n} $$ By telescoping, we see that the sequence is Cauchy, i.e. $$ a_{n+k}-a_n=\sum_{j=1}^ka_{n+j}-a_{n+j-1}\leq\sum_{j=1}^k\frac{n+j+1}{2^{n+j}}\leq\frac3{2^n}. $$ So the limit $L$ exists.

The quantity $n^{2^{-n}}$ tends to $1$ as $n \to \infty$, so we can find a constant $C$ such that

$$ n^{2^{-n}} \leq C $$

for all $n$, and hence that

$$ n \leq C^{2^n} $$

for all $n$. Thus

$$ \begin{align} a_n &= \sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}} \\ &\leq \sqrt{C^2+\sqrt{C^{2^2}+\sqrt{C^{2^3}+\cdots+\sqrt{C^{2^n}}}}} \\ &= C \sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}} \\ &< C \sqrt{1+\sqrt{1+\sqrt{1+\cdots}}} \\ &= C\varphi, \end{align} $$

where $\varphi$ is the golden ratio. The sequence $a_n$ is bounded and increasing and therefore has a limit.

This essentially mimics the proof of Herschfeld's theorem, the statement of which can be found in my answer here.