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!

Solution 1:

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 $\phi$.

Solution 2:

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}$

Solution 3:

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:

$$1.27207<L<1.27228$$