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

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

• Commented Sep 19, 2013 at 17:10
• @RonGordon, great link. After posting the answer I was trying to put the exact value of the series which is given at your link. Commented Sep 19, 2013 at 17:14
• This (oeis.org/A259235) and its square root (oeis.org/A112302) can be found on OEIS. Commented May 17 at 4:08

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.

• the Stronger Induction article is interesting indeed. Just a small typo in worked out example #1, last line. It should read: $\dfrac{1}{n(n+1)} > \dfrac{1}{(n+1)^2}$. The sign over there is the other way round. Commented Sep 19, 2013 at 18:49
• @ParthThakkar Fixed. Thanks! Commented Sep 19, 2013 at 18:51
• It seems like if you only prove strict inequality when you fix $n$, then you might get equality when you take the limit $n \to \infty$. However that's a minor point; if we have a constant $C$ that gives weak inequality then we can just add an arbitrarily small constant to get strict inequality. Interestingly, I think that your sequence of upper bounds converges to the actual value of the expression. So this is about the best set of upper bounds one could hope for. Commented Sep 19, 2013 at 18:51
• @user2566092 Indeed, made $\leq$. It also shows you that the initial terms are really important (which easily follows by the exponent). Having control over them greatly improves the bound. In the subsequent terms, we are using $2^{\frac{1}{2}}$, $3^\frac{1}{4}$, etc explicitly. Commented Sep 19, 2013 at 18:54

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 $C<2.8434$.

• But that only proves $C \lt \exp(2)$, which is not very tight. Commented Sep 19, 2013 at 17:20
• @RossMillikan Corrected! Thanks for the comment. Commented Sep 19, 2013 at 17:29
• This looks almost right, except that $\log 3 > 1$, but not by much. So I think that if that is corrected for, a valid $C$ will be found which is hopefully still less than $3$. Currently my computer calculation shows that $C$ should be greater than $2.76$. Commented Sep 19, 2013 at 17:59
• To 50 digits: 2.7612068419574980332304546465801311048761259807153 (using Pari/GP up to 300 terms of the $\log C$-formula) Commented Sep 19, 2013 at 18:08
• $\sum_{i=1}^{+\infty}\frac{i}{2^i}=2$, so the actual proof gives only $C<\exp\left(\frac{3}{2}+\frac{1}{40}\right)=4.59514\ldots$. Commented Sep 19, 2013 at 19:25

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.

• What does $\zeta'(-q)$ look like asymptotically? Does it go to zero exponentially? Commented May 17 at 4:04
• This (oeis.org/A259235), its square root (oeis.org/A112302), and the logarithm of the latter (oeis.org/A114124) can be found on OEIS. If you have an OEIS account (and I recommend getting one), I figure it's worth adding your formula to these entries. Commented May 17 at 4:08

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 $(\heartsuit)$.

• +1 Interesting lower bounds. I think that these comments should be included in your answer. Commented Sep 19, 2013 at 20:36
• @Calvin Lin: Upper and lower bounds deriving from the standard series acceleration technique included, thank you. Commented Sep 19, 2013 at 21:16

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 $n\longrightarrow \infty$‎.

• I slightly improved formatting using \tag{1}, etc for labels. I do hope you pass number theory, but this isn't a part of the answer.
– user147263
Commented Jul 4, 2015 at 23:34

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