Determine whether $\{a_n\}$ is convergent or not, where $$a_n:=\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}.$$
At least, we can obtain$$\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}\le\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt[n]{n}}}}<2,$$which imlies $\{a_n\}$ is bounded. But is it monotonic?
On
Another interesting estimate. When $n\geqslant 2$, $$\sqrt[n]{n}<2,\ \sqrt[n]{n+1}\leqslant 2.$$ $$1\leqslant\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n-1+\sqrt[n]{n}}}}}$$ $$\leqslant\sqrt[n]{{\color{red}{(n-1)}}+\sqrt[n]{{\color{red}{(n-1)}} +\sqrt[n]{{\color{red}{(n-1)}}+\cdots+\sqrt[n]{{\color{red}{(n-1)}}+\sqrt[n]{n}}}}}$$ $$\leqslant\sqrt[n]{(n-1)+\sqrt[n]{(n-1)+\sqrt[n]{(n-1)+\cdots+\sqrt[n]{n+1}}}}$$ $$\leqslant\sqrt[n]{n+1}\to 1\ \text{as}\ n\to \infty.$$
We have $$ \sqrt[n]{n} \le n, $$ $$ \sqrt[n]{{n - 1 + \sqrt[n]{n}}} \le \sqrt[n]{{n + \sqrt[n]{n}}} \le \sqrt[n]{{2n}} \le 2n, $$ $$ \sqrt[n]{{n - 2 + \sqrt[n]{{n - 1 + \sqrt[n]{n}}}}} \le \sqrt[n]{{n + \sqrt[n]{{n - 1 + \sqrt[n]{n}}}}} \le \sqrt[n]{{n + 2n}} \le 3n, $$ etc. Hence $$ \sqrt[n]{{1 + \sqrt[n]{{2 + \cdots + \sqrt[n]{n}}}}} \le \sqrt[n]{{1 + (n - 1)n}} \to 1. $$ On the other hand $a_n >1$. Hence, the limit is $1$.