# Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$

I have this problem which says that for any positive integer $$n$$, $$n \neq 0$$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$ This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)

• Can you show us your attempt? You will be more likely to get a response. – Sam Streeter Oct 25 '18 at 15:03
• Maple gives us this term for your product: $${\frac {\cosh \left( 1/2\,\pi\,\sqrt {3} \right) \left( n+1 \right) \Gamma \left( n+1/2+i/2\sqrt {3} \right) \Gamma \left( n+1/2-i/2\sqrt {3} \right) }{\pi\, \left( \Gamma \left( n+1 \right) \right) ^{2}}}$$ – Dr. Sonnhard Graubner Oct 25 '18 at 15:09
• I suspect someone will prove the $n\to\infty$ limit is $\pi^{-1}\cosh\frac{\pi\sqrt{3}}{2}$. – J.G. Oct 25 '18 at 15:19
• @J.G. You were right. – user593746 Oct 25 '18 at 16:36
• – Sil Nov 5 '18 at 8:04

The cases $$n=1$$ and $$n=2$$ can be verified manually. We assume that $$n\geq 3$$. For an integer $$k>2$$, we have $$1+\frac{1}{k^3}=\left(1+\frac1k\right)\left(1-\frac1k+\frac1{k^2}\right)=\left(1+\frac1k\right)\left(\frac{k-1}{k}\right)^2\left(1+\frac{1}{k-1}+\frac{1}{(k-1)^2}\right).$$ We note that $$1+\frac{1}{k-1}+\frac1{(k-1)^2}=\frac{1-\frac{1}{(k-1)^3}}{1-\frac{1}{k-1}}<\frac{1}{1-\frac{1}{k-1}}=\frac{k-1}{k-2}$$ for $$k>2$$. That is, $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)\leq \left(1+\frac1{1^3}\right)\left(1+\frac1{2^3}\right)\prod_{k=3}^n\left(1+\frac1k\right)\left(\frac{k-1}{k}\right)^2\left(\frac{k-1}{k-2}\right).$$ The RHS can be telescoped nicely: $$\prod_{k=3}^n\left(1+\frac1k\right)=\frac{n+1}{3},$$ $$\prod_{k=3}^n\left(\frac{k-1}{k}\right)=\frac{2}{n},$$ and $$\prod_{k=3}^n\left(\frac{k-1}{k-2}\right)=n-1.$$ Thus, $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)\leq 2\left(\frac98\right)\left(\frac{n+1}{3}\right)\left(\frac{2}{n}\right)^2(n-1)=3\left(\frac{n^2-1}{n^2}\right)<3.$$

In fact, for a fixed positive integer $$m\geq 3$$ and for every $$n\geq m$$, we have $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)\leq t_m\left(\frac{n^2-1}{n^2}\right) where $$t_m=\frac{m^2}{m^2-1}\ \prod_{k=1}^m\left(1+\frac{1}{k^3}\right).$$ If we pick $$m=5$$, we get $$m=\frac{637}{256}<\frac{640}{256}=\frac52$$. So, we can prove a stronger inequality $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)<\frac52.$$

By writing $$1+\frac{1}{k^3}$$ as $$\frac{(k+1)(k-u)(k-v)}{k^3}=\frac{k+1}{k}\left(\frac{\Gamma(k+1-u)}{\Gamma(k-u)}\right)\left(\frac{\Gamma(k+1-v)}{\Gamma(k-v)}\right)\left(\frac{\Gamma(k)}{\Gamma(k+1)}\right)^2,$$ where $$u=\frac{1+\sqrt{3}i}2$$ and $$v=\bar{u}=1-u$$, we have $$\prod_{k=1}^n\left(1+\frac1{k^3}\right)=\frac{(n+1)\Gamma(n+1-u)\Gamma(n+1-v)}{\Gamma(1-u)\Gamma(1-v)\big(\Gamma(n+1)\big)^2}.$$ From the reflection formula, and from the relation $$u+v=1$$, we have $$\Gamma(1-u)\Gamma(1-v)=\Gamma(1-u)\Gamma(u)=\left(\frac{\pi}{\sin(\pi u)}\right).$$ Since \begin{align}\sin(\pi u)&=\sin\left(\frac{\pi}{2}+\frac{\sqrt{3}}{2}\pi i\right)\\&=\sin\left(\frac{\pi}{2}\right)\cosh\left(\frac{\sqrt{3}}{2}\pi\right)+i\cos\left(\frac{\pi}{2}\right)\sinh\left(\frac{\sqrt{3}}{2}\pi\right) \\&=\cosh\left(\frac{\sqrt{3}}{2}\pi\right),\end{align} we conclude that $$\Gamma(1-u)\Gamma(1-v)=\frac{\pi}{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}.$$ This leads to \begin{align}\prod_{k=1}^n\left(1+\frac1{k^3}\right)&=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}\left(\frac{(n+1)\Gamma(n+1-u)\Gamma(n+1-v)}{\big(\Gamma(n+1)\big)^2}\right) \\&=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}\left(\frac{\Gamma(n+1-u)\ (n+1)^u}{\Gamma(n+1)}\right)\left(\frac{\Gamma(n+1-v)\ (n+1)^v}{\Gamma(n+1)}\right).\end{align} Since $$\lim_{n\to\infty}\frac{\Gamma(n+z)}{\Gamma(n)\ n^z}=1$$ for all $$z\in\mathbb{C}$$, we get $$\prod_{k=1}^\infty\left(1+\frac1{k^3}\right)=\frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}\approx 2.42818979.$$

• Thanks that was very helpful – Lazar Ionut Radu Oct 25 '18 at 15:50
• But if you can tell me what was the thought process for factoring $1+\frac{1}{k^3}$ that way? – Lazar Ionut Radu Oct 25 '18 at 15:51
• First I factorized $1+\frac1{k^3}=\left(1+\frac1k\right)\left(1-\frac1k+\frac1{k^2}\right)$. I wanted to somehow rewrite $1-\frac1k+\frac1{k^2}$ as something of the form $1+a+a^2$ for $0<a<1$, so that I can use $$1+a+a^2<1+a+a^2+a^3+\ldots=\frac{1}{1-a}.$$ So I observed that $$1-\frac1k+\frac1{k^2}=\frac{k^2-k+1}{k^2}=\frac{(k-1)^2+(k-1)+1}{k^2}.$$ If I changed the denominator to be $(k-1)^2$, then I would get $$\frac{(k-1)^2+(k-1)+1}{(k-1)^2}=1+\frac{1}{k-1}+\frac{1}{(k-1)^2},$$ which is what I need. That is why you have the "rescaling term" $\left(\frac{k-1}{k}\right)^2$. – user593746 Oct 25 '18 at 15:54
• I wish I could upvote once for the inequality and again for the exact calculation. – saulspatz Oct 25 '18 at 16:42

Denote $$p(n):=(1+\frac{1}{1^3})(1+\frac{1}{2^3})(1+\frac{1}{3^3})...(1+\frac{1}{n^3})$$.

Claim: $$p(n)\leq3-\frac2{n^2},\,\forall n\geq2.$$

For $$n=2$$, we have $$\frac94\leq3-\frac12$$.

Then suppose $$p(n)\leq3-\frac2{n^2}$$ for some $$n$$. We see \eqalign{ p(n+1)&=p(n)(1+\frac1{(n+1)^3})\cr &\leq3-\frac{2}{n^2}+\frac3{(n+1)^3}-\frac{2}{n^2(n+1)^3}\cr &=3+\frac{3n^2-2(n^3+3n^2+3n+1)-2}{n^2(n+1)^3}\cr &=3-\frac{2n^3+3n^2+6n+4}{n^2(n+1)^3}\cr &=3-\frac{2n^3+2n^2+(n^2+6n+4)}{n^2(n+1)^3}\cr &\leq3-\frac{2n^2(n+1)}{n^2(n+1)^3}\cr &\leq3-\frac2{(n+1)^2}}.

As pointed out by @saulspatz, one can prove that $$p(n)\leq3-\frac1n,\forall n\geq1$$ by the same method.

Hope this helps.

• Thanks, this helps a lot – Lazar Ionut Radu Oct 25 '18 at 16:57
• +1 If I'm not mistaken, you can prove $p(n)\geq3-{1\over n}, n\geq1$ by the same method an a little simpler algebra. – saulspatz Oct 25 '18 at 17:11
• @saulspatz I think you mean $p(n)\leq3-\frac1n$. Indeed this is a sharper result and easier to prove. Thanks for pointing this out! – awllower Oct 26 '18 at 0:06
• @awllower Yes, that was a typo, but it's too late to edit the comment. – saulspatz Oct 26 '18 at 4:35

We have that

$$\prod_{k=1}^\infty \left(1+\frac{1}{k^3}\right)<3\iff \sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 3$$

and since $$\forall x>0\, \log(1+x)

$$\sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)=\log 2+\sum_{k=2}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 2+\sum_{k=2}^\infty \frac{1}{k^3}<\log 3$$

Taking seriously the suggestions of Dr. Sonnhard Graubner and J.G., one can indeed prove that, for $$x\in\mathbb{C}$$ and $$p\in\mathbb{N}_+$$, $$\prod_{n=1}^\infty \left(1+\frac{x^p}{n^p}\right) = \prod_{j=1}^p \frac{1}{\Gamma(1+\omega_p^j x)} ,$$ where $$\{-\omega_p^j\}_{j=1}^p$$ are the $$p$$-th roots of $$-1$$.

In fact, using that $$\sum_{j=1}^p \omega_p^j = 0$$ and this identity (see also this), one has that $$\prod_{n=1}^N \left(1+\frac{x^p}{n^p}\right) = \prod_{j=1}^p \prod_{n=1}^N \left(1+\frac{\omega_p^j x}{n}\right) = \prod_{j=1}^p e^{-\omega_p^j x \gamma} \prod_{n=1}^N \left(1+\frac{\omega_p^j x}{n}\right) e^{-\frac{\omega_p^j x}{n}}$$ converges, as $$N\to\infty$$, to $$\prod_{j=1}^p \frac{1}{\omega_p^j x \Gamma(\omega_p^j x)} = \prod_{j=1}^p \frac{1}{\Gamma(1+\omega_p^j x)}.$$

Specializing for $$p=3$$ and $$x=1$$ and using the formula for the absolute value here, we have $$\prod_{n=1}^\infty \left(1+\frac{1}{n^p}\right) = \frac1{\Gamma(1+1)\Gamma(1-\frac12-\frac{\sqrt3 i}2)\Gamma(1-\frac12+\frac{\sqrt3 i}2)} = \frac1{|\Gamma(\frac12+\frac{\sqrt3 i}2)|} = \frac{\cosh\left(\frac{\sqrt3 }2\pi\right)}{\pi} .$$

• Very nice. Indeed, if $x_1,x_2,\ldots,x_m$ and $y_1,y_2,\ldots,y_m$ are complex numbers which are not positive integers such that $x_1+x_2+\ldots+x_m=y_1+y_2+\ldots+y_m$. Then, $$\prod_{n=1}^\infty\,\prod_{j=1}^m\,\left(\frac{n-x_j}{n-y_j}\right)=\prod_{j=1}^m\,\left(\frac{\Gamma\left(1-y_j\right)}{\Gamma\left(1-x_j\right)}\right)\,.$$ – Batominovski Oct 26 '18 at 10:47
• On the other hand, if $\displaystyle\text{Re}\left(\sum_{j=1}^m\,x_j-\sum_{j=1}^m\,y_j\right)>0$, then $\displaystyle\prod_{n=1}^\infty\,\prod_{j=1}^m\,\left(\frac{n-x_j}{n-y_j}\right)=0$, and if $\displaystyle\text{Re}\left(\sum_{j=1}^m\,x_j-\sum_{j=1}^m\,y_j\right)<0$, then $\displaystyle\prod_{n=1}^\infty\,\prod_{j=1}^m\,\left(\frac{n-x_j}{n-y_j}\right)=\infty$. – Batominovski Oct 26 '18 at 10:51
• If $\displaystyle\text{Re}\left(\sum_{j=1}^m\,x_j-\sum_{j=1}^m\,y_j\right)=0$, but $\displaystyle\sum_{j=1}^m\,x_j-\sum_{j=1}^m\,y_j\neq 0$, then $\displaystyle\prod_{n=1}^\infty\,\prod_{j=1}^m\,\left(\frac{n-x_j}{n-y_j}\right)$ does not converge at all in $\mathbb{C}\cup\{\infty\}$. – Batominovski Oct 26 '18 at 10:56
• @Batominovski very interesting generalization! – Federico Oct 26 '18 at 11:25

With $$AM-GM$$ \begin{align} 1.(1+\frac{1}{2^3})(1+\frac{1}{3^3})\cdots(1+\frac{1}{n^3}) &\leq\left(\dfrac1n(n+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots+\frac{1}{n^3})\right)^n \\ &\leq\left(\dfrac1n(n+\sum_{n=1}^\infty\frac{1}{n^3}-1)\right)^n \\ &\leq\left(\dfrac1n(n+\zeta(3)-1)\right)^n \\ &\leq\left(1+\dfrac{\zeta(3)-1}{n}\right)^n \\ &< e^{\zeta(3)-1}\\ &<\frac32 \end{align} Thanks to Winther.

• That doesn't work because $\zeta(3)>\ln 3> 1$. – J.G. Oct 25 '18 at 15:12

By induction we can prove the stronger

$$\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$$

indeed

1. base cases: by inspection the inequality is satisfied for for $$n=1,2, 3$$

2. induction step:

• assume true that (Ind. Hyp.): $$\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$$

• we want to prove that: $$\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n+1}\right)$$

then we have

$$\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)=\prod_{k=1}^n\left(1+\frac{1}{k^3}\right) \cdot \left(1+\frac{1}{(n+1)^3}\right)<$$ $$\stackrel{Ind. Hyp.}<3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$

thus we need to show that

$$3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$

which is true indeed

$$1+\frac{1}{{n+1}^3}-\frac1n-\frac{1}{n(n+1)^3}\stackrel{?}<1-\frac{1}{n+1}$$

$$n-(n+1)^3-1\stackrel{?}<-n(n+1)^2$$

$$n-n^3-3n^2-3n-1-1\stackrel{?}<-n^3-2n^2-n$$

$$n^2+n+2\stackrel{?}>0$$

• Very helpful thanks – Lazar Ionut Radu Oct 25 '18 at 16:33
• @LazarIonutRadu I just want to give an answer to your original question for a method by induction, it is interesting since how you can see we need to consider a stronger inequality to can proceed by induction. It is a typical case we can encounter with the method. – gimusi Oct 25 '18 at 16:36

Claim:

For all positive integers $$n$$, we have $$\prod_{k=1}^n \left(1+\frac{1}{k^3}\right) < e$$ or equivalently $$\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) < 1$$ Proof:

It suffices to show that $$\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) < 1-\frac{1}{(n+1)^2}\tag{*}$$ holds for all positive integers $$n$$.

To prove $$(*)$$, proceed by induction on $$n$$.

By direct evaluation, $$(*)$$ holds for the base case $$n=1$$.

Suppose $$(*)$$ holds for some positive integer $$n$$. \begin{align*} \text{Then}\;\;&\sum_{k=1}^{n+1} \ln\left(1+\frac{1}{k^3}\right)\\[4pt] =\;& \left( \sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) \right) + \ln\left(1+\frac{1}{(n+1)^3}\right) \\[4pt] <\;& \left( 1-\frac{1}{(n+1)^2}\right) + \ln\left(1+\frac{1}{(n+1)^3}\right) &&\text{[by the inductive hypothesis]} \\[4pt] <\;& \left( 1-\frac{1}{(n+1)^2}\right) + \frac{1}{(n+1)^3} &&\text{[since \ln(1+x) < x, for all x > 0]} \\[4pt] =\;&\left(1-\frac{1}{(n+2)^2}\right)-\frac{n^2+n-1}{(n+1)^3(n+2)^2} \\[4pt] <\;&1-\frac{1}{(n+2)^2} &&\text{[since n^2+n-1 > 0]} \\[4pt] \end{align*} which completes the induction, and thus proves the claim.

Less Than $$\boldsymbol{3}$$

The inequality $$1+\frac1{n^3}\lt\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\tag1$$ can be verified by cross-multiplying and then multiplying both sides by $$2n^5(n-1)^2$$; that is, $$2n^7-4n^6+3n^5\underbrace{-3n^3+3n^2-2n+1}_\text{-(3n^2+2)(n-1)-1\lt0 for n\ge1}\lt2n^7-4n^6+3n^5\tag2$$ Therefore, employing a telescoping product, \begin{align} \prod_{n=1}^\infty\left(1+\frac1{n^3}\right) &\lt2\prod_{n=2}^\infty\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\\ &=2\cdot\frac32\\[9pt] &=3\tag3 \end{align}

Actual Value \begin{align} \lim_{n\to\infty}\prod_{k=1}^n\frac{k^3+1}{k^3} &=\lim_{n\to\infty}\frac{\Gamma(n+2)\,\Gamma\!\left(n+\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(n+\frac12-i\frac{\sqrt3}2\right)}{\Gamma(2)\,\Gamma\!\left(\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(\frac12-i\frac{\sqrt3}2\right)\Gamma(n+1)^3}\tag4\\ &=\frac1{\Gamma\!\left(\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(\frac12-i\frac{\sqrt3}2\right)}\\ &\times\lim_{n\to\infty}\frac{\Gamma(n+2)\,\Gamma\!\left(n+\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(n+\frac12-i\frac{\sqrt3}2\right)}{\Gamma(n+1)^3}\tag5\\ &=\frac{\sin\left(\frac\pi2+i\frac{\pi\sqrt3}2\right)}{\pi}\times1\tag6\\[6pt] &=\frac{\cosh\left(\frac{\pi\sqrt3}2\right)}{\pi}\tag7 \end{align} Explanation:
$$(4)$$: $$\prod\limits_{k=1}^n(k+x)=\frac{\Gamma(n+1+x)}{\Gamma(1+x)}$$ and $$k^3+1=(k+1)\left(k-\frac12+i\frac{\sqrt3}2\right)\left(k-\frac12-i\frac{\sqrt3}2\right)$$
$$(5)$$: pull out the constant factor using $$\Gamma(2)=1$$
$$(6)$$: apply Euler's Reflection Formula $$\Gamma(x)\,\Gamma(1-x)=\frac\pi{\sin(\pi x)}$$
$$\phantom{(6)\text{:}}$$ and Gautschi's Inequality, which implies $$\lim\limits_{n\to\infty}\frac{\Gamma(n+x)}{\Gamma(n)\,n^x}=1$$
$$(7)$$: $$\cos(ix)=\cosh(x)$$