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Question Number 90435 by naka3546 last updated on 23/Apr/20

Commented by MJS last updated on 23/Apr/20

$$\mathrm{none}\mathrm{of}\mathrm{these}\mathrm{options}$$$$\sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{...}}}}}=2^{\frac{1}{2^1}}{3}^{\frac{1}{2^2}}{4}^{\frac{1}{2^3}}{5}^{\frac{1}{2^4}}...n^{\frac{1}{2^{n-1}}}=$$$$=\underset{k=2}{\prod ^n}{k}^{\frac{1}{2^{k-1}}}={\mathrm{e}}^{\underset{k=2}{\sum ^n}\frac{1}{2^{k-1}}\ln k}={\mathrm{e}}^{2\underset{k=2}{\sum ^n}\frac{\ln k}{2^k}}$$$$\mathrm{is}\mathrm{there}\mathrm{an}\mathrm{exact}\mathrm{value}\mathrm{for}\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{\ln k}{2^k}?$$$$\mathrm{I}\mathrm{get}\approx .507833922u868$$$$\Rightarrow \mathrm{answer}\mathrm{of}\mathrm{question}\approx 2.76120684196$$

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