How to get value of $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}$?
My solution so far is to take logarithm, split logarithm as row and sum column-vise: $$\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}} = L$$ $$\ln(L)=\frac{1}{2}\ln1+\frac{1}{4}\ln2+\frac{1}{8}\ln3+\frac{1}{16}\ln4+\cdots$$ Using $$\ln(x+1)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots$$ I get: $$\ln(L)=\frac{1}{2}0-\frac{1}{2}\frac{0^{2}}{2}+\frac{1}{2}\frac{0^{3}}{3}-\frac{1}{2}\frac{0^{4}}{4}+\cdots$$ $$+\frac{1}{4}1-\frac{1}{4}\frac{1^{2}}{2}+\frac{1}{4}\frac{1^{3}}{3}-\frac{1}{4}\frac{1^{4}}{4}+\cdots$$ $$+\frac{1}{8}2-\frac{1}{8}\frac{2^{2}}{2}+\frac{1}{8}\frac{2^{3}}{3}-\frac{1}{8}\frac{2^{4}}{4}+\cdots$$ summing by columns i get: $$ =(\frac{1}{2}0+\frac{1}{4}1+\frac{1}{8}2+\frac{1}{16}3+\cdots)-(\frac{1}{2}\frac{0^{2}}{2}+\frac{1}{4}\frac{1^{2}}{2}+\frac{1}{8}\frac{2^{2}}{2}+\frac{1}{16}\frac{3^{2}}{2}+\cdots)+$$ $$+(\frac{1}{2}\frac{0^{3}}{3}+\frac{1}{4}\frac{1^{3}}{3}+\frac{1}{8}\frac{2^{3}}{3}+\frac{1}{16}\frac{3^{3}}{3}+\cdots)-(\frac{1}{2}\frac{0^{4}}{4}+\frac{1}{4}\frac{1^{4}}{4}+\frac{1}{8}\frac{2^{4}}{8}+\frac{1}{16}\frac{3^{4}}{4}+\cdots)+\cdots$$
The "only" thing left is how to figure out these summs in each bracket. Any idea hot to do that, or you may suggest another solution? Best regards.
