<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] Reusing my answer on Quora and showing how it can be used to improve bounds
Write the expression as
$$\displaystyle 2\sqrt[2]{\frac{3}{2}}\sqrt[4]{\frac{4}{3}}\sqrt[8]{\frac{5}{4}}\sqrt[16]{\frac{6}{5}}...$$
$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{1}{2^{n-1}}}$$
$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{n}{n2^{n-1}}} < 2\prod\limits_{n=2}^{+\infty} e^{\frac{1}{n2^{n-1}}}$$
Now
$$\displaystyle \prod\limits_{n=1}^{+\infty} e^{\frac{1}{n2^{n-1}}}=4$$
because
$$\displaystyle \sum\limits_{n=1}^{+\infty} \frac{1}{n2^{n-1}}=\log(4)$$
$$\displaystyle 2\prod\limits_{n=2}^{+\infty} e^{\frac{1}{n2^{n-1}}}=2\frac{4}{e}<3$$
The expression can be used to get more convergents, simply take first few terms and set the rest to the exponent.
$$\frac{8\sqrt[4]{3}}{e^{\frac{4}{3}}}\approx 2.77$$
$$4\frac{2^\frac{3}{4}\sqrt[4]{3}\sqrt[8]{5}}{e^{\frac{131}{96}}} \approx 2.766$$ already
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