<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] Put $K=\log C$. The problem is then to found a tight upper bound for:
$$ K = \sum_{n=1}^{+\infty}\frac{\log(1+n)}{2^n}. $$
Note that:
$$ K/2=K-K/2 = \sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{n}\right)}{2^n}. $$
By writing $K/4$ as $K/2-K/4$ just like above, we get:
$$\frac{K}{4}=\frac{\log 2}{2}-\sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{n(n+2)}\right)}{2^{n+1}},$$
equivalent to:
$$(\heartsuit)\quad K = \log 3 -\sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{(n+1)(n+3)}\right)}{2^n} = \log 3 +\sum_{n=1}^{+\infty}\frac{\log\left(1-\frac{1}{(n+2)^2}\right)}{2^n}.$$
Now using the Bernoulli inequality we have:
$$\log 3-\sum_{n=1}^{+\infty}\frac{1}{(n+1)(n+3)2^n}\leq K \leq \log 3-\sum_{n=1}^{+\infty}\frac{1}{(n+2)^2 2^n}, $$
that gives:
$$ 1 < 3\log 2 + \log 3 -\frac{13}{6} \leq K \leq \log 3+\frac{1}{12}\left(27-4\pi^2+24\log^2 2\right), $$
so $K$ is between $1.0113\ldots$ and $1.0196$, and $C$ is between $2.7494\ldots$ and $2.7722\ldots$.
The next step of the acceleration process leads us to:
$$ K = 3\log 2-\log 3+\sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{5+2n}{(n+1)(n+3)^3}\right)}{2^n},$$
that converges faster but is way less appealing than $(\heartsuit)$.
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