Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$

Question

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 $n+1$.

Take the log and you want to prove:$$\sum_{k=2}^n \frac{\log k}{2^{k-1}} < \log 3$$ — Thomas Andrews, Nov 13 '14 at 22:37

Thomas Andrews · Accepted Answer · 2014-11-13T23:43:13.717

11

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.

edited Nov 13 '14 at 23:43

answered Nov 13 '14 at 22:53

Thomas Andrews

186,215

The sequence is increasing as a function of $n$, isn't it? But your upper bound is decreasing. – TonyK Nov 13 '14 at 22:58
Nice idea, original :) – Peter Franek Nov 13 '14 at 22:59
That's not a contradiction, it would just show that the sequence is always less than $2$. But that is an error by calculation. @TonyK – Thomas Andrews Nov 13 '14 at 22:59
Indeed. And in fact we already have $\sqrt{2\sqrt{3\sqrt{4}}} \approx 2.213 > 2+\frac{1}{16}$. – TonyK Nov 13 '14 at 23:01
I don't see why is it wrong or why $1/2^n + \sum_{k=1}^n k/(2^k)$ is decreasing. – Peter Franek Nov 13 '14 at 23:02
It's wrong because, as Tony says, $n=4$ is false. I'm just missing something in the computation - it might still work, but $2+\frac{1}{2^n}$ is the wrong answer. – Thomas Andrews Nov 13 '14 at 23:04
There are just $n-1$ square roots in the original expression. – Peter Franek Nov 13 '14 at 23:08
Yes, just realized, I got the wrong count. It's $n-1$, though, @Peter - think of the case $n=2$, then $\sqrt{2}$ is the value. – Thomas Andrews Nov 13 '14 at 23:09
Throw out all those superfluous $1$'s, and I think it works. – TonyK Nov 13 '14 at 23:13
1

Yes, finally fixed it, @TonyK. – Thomas Andrews Nov 13 '14 at 23:14
Now all that remains is to change one instance of "increasing" to "decreasing" :-) – TonyK Nov 13 '14 at 23:15
Yes, I'm talking about the original sequence, which is definitely increasing. Basically, $a_n$ is a strictly increasing sequence with $a_n<3+\frac{1}{2^{n-1}}$, which means $a_n<3$ for every $n$... @TonyK $\sup a_n \leq \inf (3+1/2^{n-1}) = 3$. – Thomas Andrews Nov 13 '14 at 23:19
OK. I thought "this sequence" referred to your upper bound. – TonyK Nov 13 '14 at 23:26

score 2 · Answer 2 · answered Nov 13 '14 at 23:23

2

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 $\ln(3)$.

answered Nov 13 '14 at 23:23

Adhvaitha

5,450

That last equality should be $<3/256$? – Thomas Andrews Nov 13 '14 at 23:27
1

@ThomasAndrews Fixed it by changing the left hand side. – Adhvaitha Nov 13 '14 at 23:28
Some optimization on the number $10$ can be done to get the bound as $\ln(3)$. To get the desired bound, it is enough to add only $7$ terms and carry over the rest of the terms to the infinite series. – Adhvaitha Nov 13 '14 at 23:30
Your approach definitely gives a more general ability to get better estimates and error bounds. The AM/GM used in my answer is a very blunt instrument. – Thomas Andrews Nov 13 '14 at 23:33

Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$

2 Answers2