Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$

Question

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 $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$?

Answer 1

See my solution to a similar question, in which I show how to prove it by "Stronger Induction".

It tells you to prove by induction of $k \rightarrow k-1$ that

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

Hence, applying this for $k=3$, we can show that

$$ \sqrt{3 \sqrt{4 \sqrt{5 \ldots}}} \leq 4.$$

Thus, we can improve our bound to

$$ \sqrt{ 2 \times 4 } = \sqrt{8} \approx 2.82. $$

Starting with higher values of $k$ gives us a better bound. For example, with $k=4$, we get

$$ \sqrt{2 \times \sqrt{3 \times 5 } } \approx 2.78.$$

$k=5$ gives $2.769$. This seems pretty decent for very little work. You can improve on this as much as you want, but I'd stop here.

Answer 2

Take a logarithm from the both sides and use $\log n\leq n-5$ for $n\geq 7$ and we have: $$ \log C= \log \sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} = \frac{\log 2}{2}+\frac{\log 3}{2^2}+\frac{\log 4}{2^3}+\frac{\log 5}{2^4}+...\\ \leq \frac{\log 2}{2}+...+\frac{\log 6}{2^5}+\frac{2}{2^6}+\frac{3}{2^7}+...\\ \leq 0.92 +\frac{1}{16}\sum_{i=1}^{\infty} \frac{i}{2^i}=1.045 $$ which gives $C<2.8434$.

Answer 3

There is an interesting formula hiding in this product which we now try to reveal. We seek to evaluate $$ S = \log P = 2 \times \sum_{n\ge 1} \frac{\log n}{2^n}.$$ The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{\log x}{2^x}.$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{\log x}{2^x} x^{s-1} dx.$$ Observe that $$\int_0^\infty 2^{-x} x^{s-1} dx = \frac{1}{(\log 2)^s} \Gamma(s)$$ by a straightforward substitution that turns the integral into a gamma function integral. This implies that $$g^*(s) = \left(\frac{1}{(\log 2)^s} \Gamma(s)\right)' = - \frac{\log\log 2}{(\log 2)^s} \Gamma(s) + \frac{1}{(\log 2)^s} \Gamma'(s).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by $$Q(s) = \left(-\frac{\log\log 2}{(\log 2)^s} \Gamma(s) + \frac{1}{(\log 2)^s} \Gamma'(s)\right) \zeta(s).$$ The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

We have $$\operatorname{Res}(Q(s)/x^s; s=1) = -\frac{1}{x}\frac{\log\log2+\gamma}{\log 2}.$$ From the remaining residues we get the contribution $$\sum_{q\ge 0} \operatorname{Res}(Q(s)/x^s; s=-q) = -\sum_{q\ge 0} x^q \frac{(-1)^q}{q!} (\log 2)^q \left(\zeta'(-q) - \zeta(-q)\log x\right),$$ where we have used the fact that the poles of the gamma function at $s=-q$ are simple with residue $(-1)^q/q!.$

Finally to find $S(1) = S/2$ we put $x=1$ to obtain the convergent expansion $$S(1) = - \frac{\log\log2+\gamma}{\log 2} - \sum_{q\ge 0} \frac{(-1)^q}{q!} (\log 2)^q \zeta'(-q).$$ The conclusion is that $$P = \exp\left(-2\frac{\log\log2+\gamma}{\log 2} - 2\sum_{q\ge 0} \frac{(-1)^q}{q!} (\log 2)^q \zeta'(-q)\right)$$ which is approximately $$2.761206841957498033230454646580131104876125980715304850.$$ If the goal is to find approximations to $P$ it suffices to take the initial terms of the above series. For example, taking the first five terms we get five accurate digits for $P,$ taking seven terms we get seven digits and so on, the exact formula for the number of good digits is of course more complicated.

Answer 4

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 $(\heartsuit)$.

Answer 5

Iif $a(n)=\sqrt{ 2\sqrt {3\sqrt {4...\sqrt{n}}}}$‎, ‎$$a(n)^{2}= 2\sqrt {3\sqrt {4...\sqrt{n}}} \tag{1}$$‎ ‎$$\dfrac{1}{2^{2}}.a(n)^{2}= 3\sqrt {4...\sqrt{n}} \tag{2}$$‎ ‎$$\dfrac{1}{3^{2}}.\dfrac{1}{2^{2^{2}}}.a(n)^{2^{2}}= 4\sqrt{...\sqrt{n}} \tag{3}$$‎ $$\vdots$$ ‎$$\prod_{i=0}^{n-2}\Big(\dfrac{1}{(n-i)^{2^{i}}}\Big).a(n)^{2^{n-1}}=1 \tag{n}$$‎ Then $$a(n)=(\prod _{i=0}^{n-2}{(n-i)^{2^{i}})}^{\dfrac{1}{2^{n-1}}} \tag{1}$$‎ ‎and‎ ‎$$a(n+1)=(\prod _{i=0}^{n-1}{(n+1-i)^{2^{i}})}^{\dfrac{1}{2^{n}}} \tag{2}$$‎ From $(1)$ and $2$‎, ‎$$\frac{a(n+1)}{a(n)}=(n+1)^{\frac{1}{2^{n}}}$$‎

‎Mathematica could not find the limit sequence for $n\longrightarrow \infty$‎.

Answer 6

Reusing my answer on Quora and showing how it can be used to improve bounds

Write the expression as

$$\displaystyle 2\sqrt[2]{\frac{3}{2}}\sqrt[4]{\frac{4}{3}}\sqrt[8]{\frac{5}{4}}\sqrt[16]{\frac{6}{5}}...$$

$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{1}{2^{n-1}}}$$

$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{n}{n2^{n-1}}} < 2\prod\limits_{n=2}^{+\infty} e^{\frac{1}{n2^{n-1}}}$$

Now $$\displaystyle \prod\limits_{n=1}^{+\infty} e^{\frac{1}{n2^{n-1}}}=4$$ because $$\displaystyle \sum\limits_{n=1}^{+\infty} \frac{1}{n2^{n-1}}=\log(4)$$

$$\displaystyle 2\prod\limits_{n=2}^{+\infty} e^{\frac{1}{n2^{n-1}}}=2\frac{4}{e}<3$$

The expression can be used to get more convergents, simply take first few terms and set the rest to the exponent.

$$\frac{8\sqrt[4]{3}}{e^{\frac{4}{3}}}\approx 2.77$$

$$4\frac{2^\frac{3}{4}\sqrt[4]{3}\sqrt[8]{5}}{e^{\frac{131}{96}}} \approx 2.766$$ already