Prove that the product $\prod_{k= 2}^n(1-k^{-3})$ is always greater than $1/2$.

Question

Prove that $$\prod_{k=2}^{n}\left(1-\frac{1}{k^{3}}\right)\geq \frac{1}{2},\,\forall n\geq2.$$

Intuitively it looks very true and this looks very similar to the reciprocal of the Euler product. I checked in Mathematica and it seems to converge to 0.809... At n=10,000,000 Mathematica returned 0.809... Thanks in advance!

Answer 1

It is sufficient to prove the claim as $n\rightarrow \infty.$ We have $$\prod_{k\geq2}\left(1-\frac{1}{k^{3}}\right)\geq\prod_{k\geq2}\left(1-\frac{1}{k^{2}}\right) $$ now since from the Weierstrass product of the sine function $$\frac{\sin\left(\pi z\right)}{\pi z}=\prod_{k\geq1}\left(1-\frac{z^{2}}{k^{2}}\right)\Rightarrow\frac{\sin\left(\pi z\right)}{\pi z\left(1-z^{2}\right)}=\prod_{k\geq2}\left(1-\frac{z^{2}}{k^{2}}\right)\tag{1} $$ we have $$\prod_{k\geq2}\left(1-\frac{1}{k^{3}}\right)\geq\lim_{z\rightarrow1}\frac{\sin\left(\pi z\right)}{\pi z\left(1-z^{2}\right)}=\frac{1}{2} $$ as wanted. We can also prove the closed form of this product. Note that, from the Weierstass product of the hyperbolic cosine, that $$ \cosh\left(\frac{\sqrt{3}\pi}{2}\right)=\prod_{k\geq1}\left(1+\frac{3}{\left(2k-1\right)^{2}}\right)=\prod_{k\geq2}\frac{4k^{2}-12k+12}{\left(2k-3\right)^{2}} $$ so we have $$\begin{align}\prod_{k\geq2}\left(1-\frac{1}{k^{3}}\right)= & \prod_{k\geq2}\frac{\left(k-1\right)\left(k^{2}+k+1\right)}{k^{3}} \\ = & \prod_{k\geq2}\frac{k-1}{k}\prod_{k\geq2}\frac{4k^{2}+4k+4}{4k^{2}} \\ = & \prod_{k\geq2}\frac{k-1}{k}\prod_{k\geq2}\frac{\left(2k-3\right)^{2}\left(2k+3\right)}{4k^{2}\left(2k+3\right)}\prod_{k\geq2}\frac{4k^{2}+4k+4}{\left(2k-3\right)^{2}} \\ = & \prod_{k\geq2}\frac{k-1}{k}\prod_{k\geq2}\frac{2k-3}{2k+3}\prod_{k\geq2}\frac{4k^{2}-9}{4k^{2}}\prod_{k\geq2}\frac{k^{2}+k+1}{k^{2}-3k+3}\prod_{k\geq2}\frac{4k^{2}-12k+12}{\left(2k-3\right)^{2}}. \end{align} $$ Now it is not difficult to see (we have telescoping products) that $$\prod_{k=2}^{N}\frac{k-1}{k}=\frac{1}{N} $$ $$\prod_{k=2}^{N}\frac{2k-3}{2k+3}=\frac{15}{8N^{3}+12N^{2}-2N-3} $$ and $$\prod_{k=2}^{N}\frac{k^{2}+k+1}{k^{2}-3k+3}=\prod_{k=2}^{N}\frac{(k+1)^{2}-k}{(k-1)^{2}-k+2}=\frac{N^{4}+N^{2}+1}{3} $$ hence $$\lim_{N\rightarrow\infty}\prod_{k=2}^{N}\frac{k-1}{k}\prod_{k=2}^{N}\frac{2k-3}{2k+3}\prod_{k=2}^{N}\frac{k^{2}+k+1}{k^{2}-3k+3}=\frac{15}{24}. $$ Using again $(1)$ we have $$\prod_{k\geq2}\frac{4k^{2}-9}{4k^{2}}=\prod_{k\geq2}\left(1-\frac{9}{4k^{2}}\right)=\frac{\sin\left(\frac{3\pi}{2}\right)}{\frac{3\pi}{2}\left(1-\frac{9}{4}\right)}=\frac{8}{15\pi} $$ so finally $$\prod_{k\geq2}\left(1-\frac{1}{k^{3}}\right)=\color{red}{\frac{1}{3\pi}\cosh\left(\frac{\sqrt{3}\pi}{2}\right)}\approx0.809 $$ as wanted.

Answer 2

For every $k \ge 2$, $k^3 - 1 > k^3 - k$. Substitute factor by factor in your product; it becomes a telescoping product, where all factors except the first and last cancel out.

The resulting product has the limit 1/2 - which means the original product has a limit strictly greater than 1/2. A quick computation in the Oracle database shows values above 0.8 for $n$ up to 1,000,000 - not sure what you entered in Mathematica but that seems wrong. EDIT: This last comment is addressed to the OP, who originally claimed he found a limit or 1/2 in Mathematica (he has since changed his original post).

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\prod_{k=2}^{n}\pars{1 - {1 \over k^{3}}} \geq \half\,,\quad \forall\ n \geq 2:\ ?}$

---

The roots of $\ds{z^{2} + z + 1 = 0}$ are given by $\ds{r}$ and $\ds{\ol{r}}$ where $\ds{r \equiv -\,\half + {\root{3} \over 2}\,\ic}$ and $\ds{r \equiv -\,\half - {\root{3} \over 2}\,\ic}$. Nothe that $\ds{r + \ol{r} = -1}$

\begin{align} \color{#f00}{\prod_{k=2}^{n}\pars{1 - {1 \over k^{3}}}} & = \prod_{k=2}^{n}{k^{3} - 1 \over k^{3}} = \prod_{k=2}^{n}{\pars{k - 1}\pars{k - r}\pars{k - \ol{r}} \over k^{3}} \\[5mm] & =\ \overbrace{\pars{\prod_{k = 2}^{n}{1 \over k}}^{2}} ^{\ds{1 \over \pars{n!}^{2}}}\,\,\, \overbrace{\pars{\prod_{k = 2}^{n}{k - 1 \over k}}}^{\ds{1 \over n}}\,\,\, {\prod_{k = 0}^{n}\,\,\pars{k - r} \over -r\pars{1 - r}}\,\,\, {\prod_{k = 0}^{n}\,\,\pars{k - \ol{r}} \over -\ol{r}\pars{1 - \ol{r}}} \\[5mm] & = {1 \over n\pars{n!}^{\, 2}}\, \verts{{\prod_{k = 0}^{n}\,\,\pars{k - r} \over -r\pars{1 - r}}}^{2} = {1 \over n\pars{n!}^{\, 2}}\, {\verts{\vphantom{\Large A}\pars{-r}_{n + 1}}^{2} \over 3}\tag{1} \end{align} $\ds{\pars{a}_{m}\,\,\,}$ is a Pochhammer Symbol.

---

We'll use the following properties: \begin{equation} \left\lbrace\begin{array}{rcll} \ds{\pars{a}_{m}} & \ds{=} & \ds{{\Gamma\pars{a + m} \over \Gamma\pars{a}}\,,\quad} & \ds{\Gamma}\mbox{: Gamma Function} \\[2mm] \ds{\verts{\Gamma\pars{\half + y\ic}}^{2}} & \ds{=} & \ds{{\pi \over \cosh\pars{\pi y}}\,,\quad} & \ds{y \in \mathbb{R}} \\ &&& \end{array}\right. \end{equation}

---

\begin{align} \verts{\pars{-r}_{n + 1}}^{2} & = \verts{\Gamma\pars{-r + n + 1} \over \Gamma\pars{-r}}^{\, 2} = {\verts{\vphantom{\large A}\pars{n - r}!}^{\, 2} \over \verts{\vphantom{\Large A}\Gamma\pars{1/2 - \root{3}\ic/2}}^{\, 2}} \\[5mm] & = {1 \over \pi}\,\cosh\pars{{\root{3} \over 2}\,\pi} \verts{\vphantom{\large A}\pars{n - r}!}^{\, 2}\tag{2} \end{align}

---

With $\ds{\pars{1}\ \mbox{and}\ \pars{2}}$: \begin{align} \color{#f00}{\prod_{k=2}^{\infty}\pars{1 - {1 \over k^{3}}}} & = {1 \over 3\pi}\,\cosh\pars{{\root{3} \over 2}\,\pi}\,\ \overbrace{% \lim_{n \to \infty}\,\,\verts{\pars{n - r}! \over n^{1/2}\,\, n!}^{\, 2}} ^{\ds{=\ \color{#f00}{1}}}\qquad \pars{~\mbox{see}\ \ul{proof}\ \mbox{at the very end}~} \\[5mm] & = \color{#f00}{{1 \over 3\pi}\,\cosh\pars{{\root{3} \over 2}\,\pi}} \approx 0.8094 \end{align}

\begin{align}\fbox{$\ds{\ \lim_{n \to \infty}\,\,\verts{\pars{n - r}! \over n^{1/2}\,\, n!}^{\, 2}\ }$} & = \lim_{n \to \infty}\,\,\verts{{1 \over n^{1/2}}\, {\root{2\pi}\pars{n - r}^{n - r + 1/2}\,\,\expo{-\pars{n - r}} \over \root{2\pi}n^{n + 1/2}\,\,\expo{-n}}}^{\, 2} \\[5mm] & = \lim_{n \to \infty}\,\,\verts{ {n^{1/2 - \root{3}\ic/2} \over n^{1/2}}\,\,\pars{1 - {r \over n}}^{-r} \pars{1 - {r \over n}}^{n + 1/2}\,\,\expo{r}}^{\, 2} \\[5mm] & = \lim_{n \to \infty}\,\,\bracks{n^{-\root{3}\ic/2}\,\,\, n^{\root{3}\ic/2}\,\, \times 1^{2} \times \expo{-r}\,\expo{r} \expo{-\ol{r}}\,\expo{\ol{r}}} = \fbox{$\ds{\ 1\ }$} \end{align}

Answer 4

This can be proved by induction if we strengthened the inequality.

We can't induct directly since $ \frac{1}{2} ( 1 - \frac{1}{(k+1)^3 }) < \frac{1}{2}$.

However, the trick is to find a function $ g(k) $ such that

• $ g(k) ( 1 - \frac{1}{(k+1)^2 }) > g(k+1)$,
• $ g(2) < 1 - \frac{1}{2^3} $, and
• $ g(k) > \frac{1}{2}$.

With such a $g(k)$, it's clear that induction on $ \prod ( 1 - \frac{1}{k^3 }) > g(k) > \frac{1}{2}$ works.

There are several candidates for such a function. If you can't one, you can reveal it below.

$g(k) = \frac{1}{2} + \frac{1}{ k+1} $.