A question about limit Can anyone help this questions?
Find the limit of
$$1+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\sqrt[5]{5+....+\sqrt[n]{n}}}}}$$
I can only solve that this formula is less than 3 but can not find the exact answer for this.

It can be shown that $1.9<a_0<2$:
We use following inequalities:
$\sqrt[k]{k+1}>1; \sqrt[k]{k}<\sqrt[k]{k+2}<2$
Seris of $a_n$ is increasing and from first inequality we have:
$a_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n]n}}<\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n-1]{n-1+2}}}<\cdot\cdot\cdot<\sqrt{2+\sqrt[3]5}$
Therefore the series has a limit like $a_n $ such that :
$a_0\leq\sqrt{2+\sqrt[3]5}<2$
Now using first inequality we find:
$a_n>\sqrt{2+\sqrt[3]{3+\cdot\cdot\cdot+\sqrt[n-1]n}}>\cdot\cdot\cdot>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$
Therefore:
$a_0>\sqrt{2+\sqrt[3]3+\sqrt[4]5}$
It can easily be seen rhat:
$\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}>1,9$
Therefore:
$1.9<a_0<2$
