TITLE: Prove that: $\sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}}<3,\,\forall n\in\mathbb N.$
QUESTION [9 upvotes]: I know that:
$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3,$ which is one of Ramanujan's infinite radicals. So surely the expression in question is less than $3.$ But how can I prove this without mentioning this or in general how to prove:
$\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots\infty}}}}<3$ ?
I'm not quite sure, how to approach this? Expressing the expression as an infinite product: $$\prod_{i=1}^{n} i^{\frac1{2^{i-1}}},\text{ as }n\to\infty$$ and then using some sort underlying inequalities might help! Please suggest. Thanks in advance.

REPLY [4 votes]: We have $$k+1\leq {4\over3}\left({3\over2}\right)^k\qquad\bigl(k\in{\mathbb N}_{\geq1}\bigr)\ ,\tag{1}$$
with $<$ when $k\geq3$. It follows that
$$(k+1)^{1/2^k}\leq\left({4\over3}\right)^{1/2^k}\ \left({3\over2}\right)^{k/2^k}\qquad(k\geq1)\ ,$$
so that your limit $a$ satisfies
$$a=\prod_{k=1}^\infty(k+1)^{1/2^k}<{4\over3}\left({3\over2}\right)^2=3\ .$$