Prove that: $\sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}}<3,\,\forall n\in\mathbb N.$ 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.
 A: Proof without induction. Notice that 
$$\sum_{k=2}^{\infty}\frac{1}{2^{k-1}}=1\quad\mbox{and}\quad
\sum_{k=2}^{\infty}\frac{k}{2^{k-1}}=3.$$
Hence, by the concavity of the logarithm, for any integer $n\geq 2$,
$$
\begin{align}
\ln\left(\sqrt{2\sqrt{3\sqrt{4...\sqrt{n}}}}\right)&=
\ln\left(\prod_{k=2}^n k^{1/2^{k-1}}\right)< \ln\left(\prod_{k=2}^{\infty} k^{1/2^{k-1}}\right)\\&=\sum_{k=2}^{\infty}\frac{\ln(k)}{2^{k-1}}\leq
\ln\left(\sum_{k=2}^{\infty}\frac{k}{2^{k-1}}\right)=\ln(3).
\end{align}$$
A: 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
$$a=\prod_{k=1}^\infty(k+1)^{1/2^k}<{4\over3}\left({3\over2}\right)^2=3\ .$$
A: This is actually a 'standard' induction question, whose goal is to make you think about the induction hypothesis.
This is tricky because the induction is not obvious. You likely have tried applying it directly, but since
$$ \sqrt{ 2 \sqrt{3 \sqrt{\ldots \sqrt{n} } } } < \sqrt{ 2 \sqrt{3 \sqrt{\ldots \sqrt{n \sqrt{n+1}} } } }, $$
the proof fails (as seen by all the other deleted solutions).
However, this is the statement that you should induct on:

 Fix $n\geq 2$. For all values of $2\leq k \leq n$,  $\sqrt{ k \sqrt{(k+1) \sqrt{\ldots \sqrt{n} } } } < k+1 .$

Perform the 'induction' on k, going from $k$ to $k-1$ (as opposed to the typical induction on $n$ going from $n$ to $n+1$).
Specifically, the base case is when $k=n$. This is immediately obvious.
For the induction step, assume it is true for some $k$. Consider $k-1$. This induction is then immediately obvious since $(k-1)(k+1) < k^2$.
Of course, we now get a lot of other similar, interesting inequalities for free.
Moral: Choosing the correct induction hypothesis is extremely important.

Note: I personally call this method Stronger Induction (not a standard term in the literature). It cleverly choses the induction hypothesis based on observations, and includes strengthing (and modifying) the induction hypothesis like what Andre did. You can click on the link for a writeup that I did.
A: $$2^{\frac{1}{2}}3^{\frac{1}{4}}4^{\frac{1}{8}}5^{\frac{1}{16}}\cdots = \exp\sum_{n\geq 1}\frac{\log(n+1)}{2^n}=\exp S, $$
$$ 2S = \sum_{n\geq 1}\frac{\log(n+1)}{2^{n-1}}=\sum_{n\geq 0}\frac{\log(n+2)}{2^{n}}=\log(2)+\sum_{n\geq 1}\frac{\log(n+2)}{2^n},$$
$$ S = \log(2)+\sum_{n\geq 1}\frac{\log\left(1+\frac{1}{n+1}\right)}{2^n}<\log(2)+\sum_{n\geq 1}\frac{1}{(n+1)2^n}=3\log(2)-1 $$
lead to the improved bound
$$\sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{\ldots}}}}}<\color{red}{\frac{8}{e}}=2.943\ldots$$
By exploiting $\log(1+x)\leq x-\frac{x^2}{3}$ over $\left[0,\frac{1}{2}\right]$ we get the further improvement LHS $<2.787$.
A: We want control over the size of
$$a_n=2^{1/2}3^{1/4}4^{1/8}\cdots n^{1/2^{n-1}}.$$
It is convenient to take the logarithm, and show that $\log a_n\lt \log 3$. But one could also work directly with the product.
As is essentially necessary with induction proofs of inequalities, we  prove the stronger result
$$\log a_n=\frac{1}{2}\log 2+\frac{1}{4}\log 3 +\frac{1}{8}\log 4+\cdots +\frac{1}{2^{n-1}}\log n \lt \log 3 -\frac{1}{2^{n-2}}\log n.\tag{1}$$
The case $n=2$ is no problem, it comes down to the fact that $\frac{3}{2}\log 2\lt \log 3$. The inequality does hold, though not by much.
For the induction step, suppose we know that (1) holds for a particular $n$. It will then be enough to show that
$$\log 3 -\frac{1}{2^{n-2}}\log n+\frac{1}{2^n}\log(n+1)\lt \log 3 -\frac{1}{2^{n-1}}\log (n+1).$$
Some manipulation makes this inequality obvious.
