Find $\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$ $$\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$$
I have no idea about this.
The equation can be written in its recursive form as:
$$f(n) = g(1,n)$$
Where
$$g(x,n) = [x\impliedby n]\cdot (x+ g(x+1,n))^{\frac 1x}+[x=n]\cdot (n)^{\frac 1n}$$
Of course, [] is the indicator function representing of piece wise notation.
 A: 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.
A: 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:
$2.9<a_1+1<3$
