Prove $\prod_{k=1}^{2015} (1+\frac{1}{(2k+1)^3})< \frac{\sqrt{5}}{2}$

[anemone]

Prove that \(\displaystyle \prod_{k=1}^{2015}\left(1+\dfrac{1}{(2k+1)^3}\right)\lt \dfrac{\sqrt{5}}{2}.\)

 

[topsquark]

How is this an infinite product?

-Dan

 

[anemone]

Hi Dan!

Spoiler

Ops! I'm sorry! When I was creating the thread, my mind was struggling about if I should make the problem an infinite product or like the one I stated in the thread, as both are less than $\dfrac{\sqrt{5}}{2}$.

And when I finally made up my mind to make it a finite product, I forgot to change the title of the thread accordingly.

I'll re-title the thread to correct the wrong.

 

[anemone]

Prove that \(\displaystyle \prod_{k=1}^{2015}\left(1+\dfrac{1}{(2k+1)^3}\right)\lt \dfrac{\sqrt{5}}{2}.\)

Hint:

Spoiler

Note that for $x>1$, $\ln x<x-1$ always hold.

 

[anemone]

Hint #1:

Spoiler

Note that for $x>1$, $\ln x<x-1$ always hold.

Hint #2:

Spoiler

Apéry's constant $1+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\cdots=1.2020569...$ is a handy tool in proving the desired inequality \(\displaystyle \prod_{k=1}^{2015}\left(1+\dfrac{1}{(2k+1)^3}\right)\lt \dfrac{\sqrt{5}}{2}\).

 

[anemone]

My solution:

Spoiler

Note that for $x>1$, we have $\ln x<x-1$, replacing $x$ by $x+1$ we get $\ln(1+x)<x$.

So we have

$\ln\left(1+\dfrac{1}{3^3}\right)<\dfrac{1}{3^3}$

$\ln\left(1+\dfrac{1}{5^3}\right)<\dfrac{1}{5^3}$

$\ln\left(1+\dfrac{1}{7^3}\right)<\dfrac{1}{7^3}$

$\ln\left(1+\dfrac{1}{9^3}\right)<\dfrac{1}{9^3}$

$\,\,\,\,\,\,\,\,\,\,\,\,\vdots$

$\ln\left(1+\dfrac{1}{4031^3}\right)<\dfrac{1}{4031^3}$

Adding them all up gives

$\ln\left(1+\dfrac{1}{3^3}\right)+\ln\left(1+\dfrac{1}{5^3}\right)+\cdots+\ln\left(1+\dfrac{1}{4031^3}\right)<\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots+\dfrac{1}{4031^3}$

Convert the sum of natural logarithms into the natural logarithm of a product, we get

$\ln\left(\left(1+\dfrac{1}{3^3}\right)\left(1+\dfrac{1}{5^3}\right)\cdots\left(1+\dfrac{1}{4031^3}\right)\right)<\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots+\dfrac{1}{4031^3}=$

Apéry's constant tells us

\(\displaystyle \lim_{{n}\to{\infty}}\left(1+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\cdots+\dfrac{1}{n^3}\right)=\zeta (3)\)

$1+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+\dfrac{1}{6^3}+\cdots =\zeta (3)$

$\left(1+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots\right)+\left(\dfrac{1}{2^3}+\dfrac{1}{4^3}+\dfrac{1}{6^3}\cdots\right)=\zeta (3)$

$\left(1+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots\right)+\dfrac{1}{2^3}\left(1+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\cdots\right)=\zeta (3)$

$\left(1+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots\right)+\dfrac{1}{2^3}\zeta(3)=\zeta (3)$

$\left(1+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots\right)=\zeta (3)\left(1-\dfrac{1}{2^3}\right)=\dfrac{7\zeta(3)}{8}$

$\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots=\dfrac{7\zeta(3)}{8}-1$

Therefore we get

$\begin{align*}\ln\left(\left(1+\dfrac{1}{3^3}\right)\left(1+\dfrac{1}{5^3}\right)\cdots\left(1+\dfrac{1}{4031^3}\right)\right)&<\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots+\dfrac{1}{4031^3}\\&<\dfrac{1}{3^3}+\dfrac{1}{5^3}+\cdots=\dfrac{7\zeta(3)}{8}-1\end{align*}$

$\left(1+\dfrac{1}{3^3}\right)\left(1+\dfrac{1}{5^3}\right)\cdots\left(1+\dfrac{1}{4031^3}\right)<e^{\frac{7\zeta(3)}{8}-1}=e^{\frac{7(1.202)}{8}-1}\approx 1.053<\dfrac{\sqrt{5}}{2}$ and we are hence done.