Proving that $ \prod\limits_{k=1}^{n}{\left(1+\frac{1}{k^{3}}\right)}<\mathrm{e} $

Question

How would you prove that $$ \left(\forall n\in\mathbb{N}\right),\ \prod_{k=1}^{n}{\left(1+\frac{1}{k^{3}}\right)}<\mathrm{e} $$

Wolfram|Alpha says its limit would be exactly $ \frac{\cosh{\left(\frac{\sqrt{3}\pi}{2}\right)}}{\pi} $, but I think it might be nicer to solve the problem without trying to compute the infinite product. Any suggestions ?

Answer 1

Equivalently, you want $$ \sum_{k=1}^n \log \left( 1+\frac{1}{k^3}\right) < 1 $$ Now $\log(1+t) < t$, but $\sum_{k=1}^\infty 1/k^3 = \zeta(3) > 1$, so that doesn't quite do it. However, leave the $k=1$ term as $\log(2)$ and it does work:

$$ \sum_{k=1}^n \log\left(1 + \frac{1}{k^3}\right) \le \log(2) + \zeta(3) - 1 \approx 0.895204084$$

Or if you don't want to have to evaluate $\zeta(3)$ numerically, use $$ \zeta(3) - 1 = \sum_{k=2}^\infty \frac{1}{k^3} \le \int_{3/2}^\infty \frac{dx}{x^3} = \frac{2}{9} $$ where by Jensen's inequality and convexity of $1/x^3$, $$\dfrac{1}{k^3} \le \int_{k-1/2}^{k+1/2} \dfrac{dx}{x^3}$$ Then our bound is $$ \sum_{k=1}^n \log \left(1+\frac{1}{k^3}\right) \le \log(2) + \frac{2}{9} \approx 0.9153694028 $$

Answer 2

With AM>GM render for all $k\ge 2$:

$1+\dfrac{1}{k^3}<\sqrt{\dfrac{1+\dfrac{1}{k^3}}{1-\dfrac{1}{k^3}}}=\sqrt{\dfrac{k^3+1}{k^3-1}}$

We then have

$\prod_{k=2}^\infty\sqrt{\dfrac{k^3+1}{k^3-1}}=\prod_{k=2}^\infty\sqrt{\dfrac{(k+1)((k-1)^2+(k-1)+1)}{(k-1)(k^2+k+1)}}\overset{telescopes}{=}\sqrt{3/2}$

So multiplying in the $k=1$ term which equals $2$:

$\prod_{k=1}^\infty(1+\dfrac{1}{k^3})<2\prod_{k=2}^\infty\sqrt{\dfrac{k^3+1}{k^3-1}}=\sqrt{6}$

To four significant digits $\sqrt6=2.449$ and the exact value quoted by the OP $=2.428$.

It remains to prove that $e>\sqrt6$. Here we simply use the series $e=1+(1/1!)+(1/2!)+...$ (all positive terms) $>5/2>\sqrt6$.

Answer 3

You can also use the fact that $$\ln(1+x)\leq x-\frac{x^2}{3}, x\in\left[0,\frac{1}{2}\right]$$ to conclude $$\prod\limits_{k=1}^n\left(1+\frac{1}{k^3}\right)\leq2\prod\limits_{k=2}^n e^{\frac{1}{k^3}-\frac{1}{3k^6}}= 2\cdot e^{\sum\limits_{k=2}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)}$$ but $$\sum\limits_{k=2}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)= \frac{23}{192}+\sum\limits_{k=3}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)\leq \\ \frac{23}{192}+\int\limits_{2}^{n}\left(\frac{1}{x^3}-\frac{1}{3x^6}\right)dx= \frac{23}{192}+\frac{59}{480}-\frac{15 n^3-2}{30 n^5}$$ thus $$\prod\limits_{k=1}^n\left(1+\frac{1}{k^3}\right)<2\cdot e^{\frac{23}{192}+\frac{59}{480}}=\color{red}{2}\cdot e^{\frac{233}{960}}< e^{\color{red}{\frac{3}{4}}+\frac{233}{960}}=e^{\frac{953}{960}}<e$$