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!

Actually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of  the size. 
First we notice that:
$$\lim_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left(1- \left(1- \frac{1}{n} \right) \right)^{\frac{1}{n}}=1$$
This is not a proof, but the fact is well known. Now let's consider the following:
$$1< \left(\frac{1}{n-1}+1 \right)^{\frac{1}{n-1}}<1+\frac{1}{(n-1)^2}$$
$$1< \left(\frac{1}{n-2}+1+\frac{1}{(n-1)^2} \right)^{\frac{1}{n-2}}<1+\frac{1}{(n-2)^2}+\frac{1}{(n-2)(n-1)^2}$$
On the next step we get:
$$\dots 1+\frac{1}{(n-3)^2}+\frac{1}{(n-3)(n-2)^2}+\frac{1}{(n-3)(n-2)(n-1)^2}$$
In the end we obtain the following inequality:

$$\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots}}<1+\sum^{\infty}_{k=2} \frac{1}{k ~k!}=Ei(1)-\gamma=1.3179$$

We can increase precision by moving the truncated series under the radical (and we should not forget to get rid of $2$ in every denominator):
$$1+2\sum^{\infty}_{k=3} \frac{1}{k ~k!}=1+2(Ei(1)-\gamma-1-1/4)=1.135804$$

$$L < \sqrt{\frac{1}{2}+1.135804}=1.27899$$

$$1+2\cdot 3 \sum^{\infty}_{k=4} \frac{1}{k ~k!}=1+6(Ei(1)-\gamma-1-1/4-1/18)=1.074080$$

$$L < \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.074080}}=1.27305$$


Now for the lower boundary the better estimation would be:

$$L > \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1}}=1.26517$$

This is not ideal, but much more accurate than just truncating the sequence.
