Find $\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$

Summarizing the above comments: It is clear that $$ x_n = 1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}} $$ is monotonically increasing, so that the convergence of this sequence is equivalent to its boundedness.

It has been demonstrated here that the sequence $$ y_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + …+\sqrt{n}}}} $$ is convergent. Since $x_n \le y_n^2$, this implies the convergence of $(x_n)$ as well.

The exact limit of $(y_n)$ is unknown, so that one can assume that the exact limit of $(x_n)$ is difficult to compute as well.

Here I show that $1.9<Lim_{n→∞} a_n<2$ if:

$a_n=\sqrt{2+\sqrt [3]{3+\sqrt[4]{4+ . . . +\sqrt[n]{n}}}}$

We use following inequalities:

$\sqrt[k]{k+1}>1$ ; $\sqrt[k]k<\sqrt[k]{k+2}$. . (for $k>2$)

The first one is clear and the second one can be proved by induction. The series of numbers of $a_n $is increasing; using inequalities we may write:

$a_n=\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n]{n}}}<\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n-1]{n-1+2}}}< . . .<\sqrt{2+\sqrt[3]{5}}$

Therefore aeries $a_n $ has a limit equal to $a_0$ such that we have:

$a_0<\sqrt{2+\sqrt[3]{5}}<2$

Using inequalities we have:

$a_n>\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n-1]{n}}}>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$

Therefore:

$a_0>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$

We can easily see that:

$\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}>1.9$

That finally gives:

$1,9<a_n<2$

and for you question:

$2.9<a_1+1<3$