How to prove that $\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{\cdots}}}}<\sqrt[3] 2$ It's an estimation that I find interesting :
$$\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{1+\frac{1}{5}\sqrt{\cdots}}}}}<\sqrt[\leftroot{-0}\uproot{0}3]{2}$$
I think to evaluate this we need first to calculate the first ones terms . 
In fact we have : 
$$\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{1+\frac{1}{5}\sqrt{1+\frac{1}{6}}}}}}\approx 1.259611$$
And :
$$\sqrt[\leftroot{-0}\uproot{0}3]{2}\approx 1.259921$$
But I can't prove that this first decimals of the infinite nested radical are fixed .
Maybe we can use series expansion to prove that with a sequence . 
Thanks a lot for your help and your time .

For $n\in\mathbb N$, let $$a_n = \frac1n\sqrt{1+\tfrac1{n+1}\sqrt{1+\tfrac1{n+2}\sqrt{\dots}}}$$
We are interested in $a_1$. Notice that $$a_1=\sqrt{1+a_2}=\sqrt{1+\tfrac12\sqrt{1+a_3}}=\dots$$
We have $$a_n<\frac1n\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\dots}}}}=\frac\phi n,$$ where $\phi=\frac{1+\sqrt 5}2$ is the golden ratio.
Hence, $$a_1<\sqrt{1+\frac{1}{2} \sqrt{1+\frac{1}{3} \sqrt{1+\frac{1}{4} \sqrt{1+\frac{\phi}{5}
   }}}}\overset{\text{Def.}}=c$$
By numerical methods, we can guarantee that $$0.00016183\geq \sqrt[3]2-c\geq 0.00016182$$ and thus that $\sqrt[3]2> c>a_1$, which achieves the claim.
