Find the value of $$\sqrt{1\sqrt{2\sqrt {3 \sqrt{4\sqrt{5\sqrt{6\cdots\sqrt{\infty}}}}}}}$$
What is the absolute value of the root in below question and what does it represent geometrically, I had a few approaches leading to possible values to approximation and know the answers, but I also need what it represents geometrically.
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As indicated by Ron Gordon the logarithm of this is : $$l= \sum_{k=1}^\infty \frac{\log{k}}{2^k}$$ Consider the polylogarithm function : $$f(s):=\sum_{k=1}^\infty \frac{(1/2)^k}{k^s}=\operatorname{Li}_s\left(\frac 12\right)$$ then $$f'(s)=-\sum_{k=1}^\infty \frac{\log(k)}{2^k\;k^s}$$ so that the final answer may be written as $\ \boxed{\displaystyle e^{-\operatorname{Li}_{0'}\left(1/2\right)}}$ with the meaning : $e^{-\lim_{s\to 0^+}\frac d{ds}\operatorname{Li}_s\left(1/2\right)}$
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The number in question is called Somos's Quadratic Recurrence Constant, whose approximate value can be found here. It is similar to the Nested Radical Constant.
Let the limiting value of the above expression be $L$; then
$$\log{L} = \sum_{k=1}^{\infty} \frac{\log{k}}{2^k}$$