A question about limit

Question

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.

Answer 1

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:

$1.9<a_0<2$