<s>[INST] Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that  $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$? 
 [/INST] Take a logarithm from the both sides and use $\log n\leq n-5$ for $n\geq 7$ and we have:
$$
\log C= \log \sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} = \frac{\log 2}{2}+\frac{\log 3}{2^2}+\frac{\log 4}{2^3}+\frac{\log 5}{2^4}+...\\
\leq \frac{\log 2}{2}+...+\frac{\log 6}{2^5}+\frac{2}{2^6}+\frac{3}{2^7}+...\\
\leq 0.92 +\frac{1}{16}\sum_{i=1}^{\infty} \frac{i}{2^i}=1.045
$$
which gives $C<2.8434$.
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