Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.
Thus, any help to get me out of this rift is more than welcome!

Using Aron D'souza's idea further we can get:
$$L^2-\frac{1}{2}< \sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}$$
$$\sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}=\frac{1}{2} \left(1+\sqrt{\frac{7}{3}} \right)$$
$$L<\sqrt{1+\frac{1}{2} \sqrt{\frac{7}{3}}}=1.328067$$
To find the next bound we will need to solve:
$$c^4-c-\frac{1}{4}=0$$
The exact solution is too complex to write here (see Wolframalpha), so I'll just write it numerically:
$$\sqrt[4]{\frac{1}{4}+\sqrt[4]{\frac{1}{4}+\dots}}=1.0723501510383$$
The bound will become:

$$L<\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.0723501510383}}=1.272871$$

Which is three correct digits of the numerical value of the limit.

To make my answer more complete, the exact value of $c_4$ is:
$$c=\frac{1}{2} \left( b+\sqrt{\frac{2}{b}-b^2} \right)$$
$$b=\sqrt{ \sqrt[3]{ \frac{a}{18} }-\sqrt[3]{ \frac{2}{3a} } }$$
$$a=9+\sqrt{93}$$

And solving the quintic equation:
$$c^5-c-\frac{1}{5}=0$$
We get the upper bound for the limit with four correct digits:

$$L<1.272282$$

Taking into account the corresponding lower boundary:
$$L>\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\sqrt[5]{\frac{1}{5}+\sqrt[6]{\frac{1}{6}}}}}}=1.271035$$
We see that truncating the limit gives less accurate solutions than the method in this answer.
However, truncating at $\frac{1}{7}$ we can finally get very good boundaries:

$$1.27207<L<1.27228$$

