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!
 A: This is a partial result:
The underlying sequence is increasing and upper bounded by $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=\dfrac{1+\sqrt{5}}{2}=\phi$. Thus the limit exists and it is less than  $\phi$.
A: I like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} > L$ and lets assume that it converges to some constant $c$. 
Now, we can write ,
$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c$
Squaring on both sides,
$\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c^2$
Which is nothing but,
$\frac{1}{2}+c = c^2$
Solving above quadratic expression, value of $c$ will be $\dfrac{1+\sqrt{3}}{2}$ .
Thus we get slightly improved upper bound for $L$ as $ L < \dfrac{1+\sqrt{3}}{2}$
A: 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:

$$1.27207<L<1.27228$$

A: 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.
