Blaise Pascal: Little Of Everything

It's very important to know a little of everything, as it could help us to become one of the most stand out competitors who has the most competitive edge and that we cannot be replaced easily by anyone else.

Sound too good to be true? It turns out it was. :D

If you're asked to compute the value for $1 + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dfrac{1}{4^3}\cdots$, would you realize at an instant that this series couldn't be a telescoping series by all means? And could you prove this infinite series converges and do you know this has already been proved rigorously (the original proof is complex and hard to grasp) that it takes a specified value, called Apéry's constant. This value was named for Roger Apéry (1916–1994), who in 1978 proved it to be irrational. This result is known as Apéry's theorem.

(Source taken from wikipedia: Apéry's constant)

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

where $\zeta$ is the Riemann zeta function. It has an approximate value of

$\zeta(3) = 1.202056903159594285399738161511449990764986292...  $

This is called Apéry's constant.

Now I want you to algebraically manipulate the relation to find the approximate value for:

$ \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}\cdots$

Some students might want to ask if this is a trick question because they suspect there already is a well-known formula for this. Nope, this is a legit question because this value can be easily found with using the relation $\zeta(3) = 1 + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dfrac{1}{4^3}\cdots$.

Let's not beat about the bush and get to the point:

$\begin{align*}\zeta(3)&=1 + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dfrac{1}{4^3}\cdots\\&=1+\left(\dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3}+ \dfrac{1}{8^3}\cdots\right)+ \left(\dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}+ \dfrac{1}{9^3}\cdots\right)\\&=1+\dfrac{1}{2}\left(\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3}+ \dfrac{1}{4^3}\cdots\right)+ \left(\dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}+ \dfrac{1}{9^3}\cdots\right)\\&=1+\dfrac{1}{2^3}\left(1 + \dfrac{1}{2^3} + \dfrac{1}{3^3}+ \dfrac{1}{4^3}\cdots\right)+ \left(\dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}+ \dfrac{1}{9^3}\cdots\right)\\&=1+\dfrac{1}{8}\left(\zeta(3)\right)+ \left(\dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}+ \dfrac{1}{9^3}\cdots\right)\end{align*}$

Therefore, we get:

$\dfrac{7\zeta(3)}{8}-1=\dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}+ \dfrac{1}{9^3}\cdots$

i.e.

$\begin{align*}\dfrac{1}{3^3} + \dfrac{1}{5^3} + \dfrac{1}{7^3}+ \dfrac{1}{9^3}\cdots&=\dfrac{7\zeta(3)}{8}-1\\&\approx \dfrac{1.20205690315959 }{2}-1\\&\approx 0.05179979\end{align*}$

I hope you learned something useful and valuable today and you could retain the memory of this new knowledge forever and use it when the right time has come to your own advantage to solve the appropriate problems at hand.