Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ Is it true that $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ for any positive integer $n$?
We cannot prove the statement using induction as it is, because the left-hand side is increasing while the right-hand side stays constant. So we need to modify the right-hand side to something like $3-\dfrac{1}{n}$. But it is hard to quantify how much the left-hand side increases when we go from $n$ to $n+1$.
 A: We want to show that
$$\sum_{k=1}^{\infty} \dfrac{\ln(k+1)}{2^{k}} < \ln(3)$$
We have $\ln(1+k) \leq k$ for all $k \in \mathbb{N}$. Hence, we have
$$\sum_{k=1}^{\infty} \dfrac{\ln(k+1)}{2^k} = \sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} + \sum_{k=11}^{\infty} \dfrac{\ln(k+1)}{2^k} \leq \sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} + \sum_{k=11}^{\infty} \dfrac{k}{2^k}$$
$$\sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} \approx 1.013 < 1.02$$
$$\sum_{k=11}^{\infty} \dfrac{k}{2^k} = \dfrac3{256} = 0.01171875$$
Hence, the sum is less than $1.04$, which is less than $\ln(3)$.
A: You can think of the left hand side as the geometric mean of the multi-set with $2^{n-k}$ values $k$ for $k=2,\dots, n$, plus an additional one, for a total of $2^{n-1}$ elements. The AM/GM rule then gives:
$$\frac{1}{2^{n-1}}\left(1+\sum_{k=1}^n k2^{n-k}\right) = \frac{1}{2^{n-1}} + \sum_{k=2}^n\frac{k}{2^{k-1}}$$
as a strict upper bound for your formula.
Since $\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = \frac{1}{(1-1/2)^2}= 4$, and noting we subtract the term for $k=1$, that gives an upper bound of $3+\frac{1}{2^{n-1}}$.
Since the original sequence is increasing, this means that the values are$\leq 3$ for every $n$. In particular, then, each value has to be less than $3$, since the sequence is strictly increasing.
