27th April

This result is based on observations

$\displaystyle \sum_{n=0}^∞ \dfrac{x^n}{n!} = e^{1\cdot x}$ where $1$ is the root of $x-1=0$

$\displaystyle \sum_{n=0}^∞ \dfrac{x^{2n}}{(2n)!} = \dfrac{e^x+e^{-x}}{2}$ where $1 ,-1$ are the roots of $x^2-1=0$

$\displaystyle\sum_{n=0}^∞ \dfrac{x^{3n}}{(3n)!} = \dfrac{e^x+e^{\omega x}+e^{\omega^2 x}}{3}$ where $1,\omega,\omega^2$ are the roots of $x^3-1=0$ (Proved by sir @Chris Lewis and sir@Chew-Seong Cheong in this problem )

We can also express $\displaystyle\sum_{n=0}^∞ \dfrac{x^{4n}}{(4n)!} = \dfrac{e^{ix}+e^{-ix} +e^{1\cdot x} +e^{-1\cdot x} }{4} = \dfrac{\cos(x)+\cosh(x)}{2}$ where $i,-i,1,-1$ are the roots of $x^4-1=0$

So for general $\displaystyle\sum_{n=0}^∞ \dfrac{x^{an}}{(an)!} = \dfrac{\sum_{r=1}^a e^{a_r x} }{a}$ where $a_1, a_2 ,..a_a$ are the roots of $x^a -1=0$

So $\sum_{n=0}^∞ \dfrac{x^{6n}}{(6n)!} = \dfrac{e^x +e^{-x} + e^{x/2} (e^{\frac{\sqrt{3}xi}{2}} +e^{-\frac{\sqrt{3}xi}{2}} )+e^{-x/2} (e^{\frac{\sqrt{3}xi}{2}} +e^{-\frac{\sqrt{3}xi}{2}}) }{6}$ $= \dfrac{e^x +e^{-x} +2\cos(\frac{\sqrt{3}x}{2} ) (e^{x/2} +e^{-x/2} )} {6} = \dfrac{\cosh(x)+ 2\cosh(x/2)\cos(\frac{\sqrt{3} x}{2}) }{3}$ Put $x= \dfrac{π}{\sqrt{3}}$ one will get $\sum_{n=0}^∞ \dfrac{π^{6n}}{27^n (6n)!} =\dfrac{1}{3} \cosh(\frac{π}{\sqrt{3}})= \dfrac{e^{π/\sqrt{3}} +e^{-π/\sqrt{3}}}{2\cdot 3}$

Consider the following Macluarin series of $\cosh (\omega^n x)$ , where $n=0,1,2$ and $\omega$ is the cubic root of unit. Note that $\omega = e^{\frac {2\pi}3i}$ , $\omega^3 = 1$ and $\omega^2+\omega+1 = 0$ .

$\begin{aligned} \cosh x & = \frac 1{0!} + \frac {x^2}{2!} + \frac {x^4}{4!} + \frac {x^6}{6!} + \frac {x^8}{8!} + \frac {x^{10}}{10!} + \frac {x^{12}}{12!} + \cdots \\ \cosh (\omega x) & = \frac 1{0!} + \frac {\omega^2 x^2}{2!} + \frac {\omega x^4}{4!} + \frac {x^6}{6!} + \frac {\omega^2 x^8}{8!} + \frac {\omega x^{10}}{10!} + \frac {x^{12}}{12!} + \cdots \\ \cosh (\omega^2 x) & = \frac 1{0!} + \frac {\omega x^2}{2!} + \frac {\omega^2 x^4}{4!} + \frac {x^6}{6!} + \frac {\omega x^8}{8!} + \frac {\omega^2 x^{10}}{10!} + \frac {x^{12}}{12!} + \cdots \\ \implies \cosh x + \cosh (\omega x) + \cosh (\omega^2 x) & = 3 \left(\frac 1{0!} + \frac {x^6}{6!} + \frac {x^{12}}{12!} + \frac {x^{18}}{18!} + \cdots \right) \\ \implies \frac 13 \left(\cosh \frac \pi{\sqrt 3} + \cosh \frac {\omega \pi}{\sqrt 3} + \cosh \frac {\omega^2 \pi}{\sqrt 3} \right) & = 1 + \frac {\pi^6}{27\cdot 6!} + \frac {\pi^{12}}{27^2 \cdot 12!} + \frac {\pi^{18}}{27^3\cdot 18!} + \cdots \end{aligned}$

Therefore,

$\begin{aligned} \sum_{k=0}^\infty \frac {\pi^{6k}}{27^k (6k)!} & = \frac 13 \left(\cosh \frac \pi{\sqrt 3} + \cosh \frac {\omega \pi}{\sqrt 3} + \cosh \frac {\omega^2 \pi}{\sqrt 3} \right) \\ & = \frac 13 \left(\cosh \frac \pi{\sqrt 3} + \cosh \left(\frac \pi{\sqrt 3}e^{\frac {2\pi}3i}\right) + \cosh \left(- \frac \pi{\sqrt 3}e^{\frac {2\pi}3i}\right) \right) \\ & = \frac 13 \left(\cosh \frac \pi{\sqrt 3} + \cosh \left(-\frac \pi{2\sqrt 3} + \frac \pi 2 i\right) + \cosh \left(-\frac \pi{2\sqrt 3} - \frac \pi 2 i\right) \right) \\ & = \frac 16 \left(e^{\frac \pi{\sqrt 3}} + e^{-\frac \pi{\sqrt 3}} + e^{-\frac \pi{\sqrt 3}+\frac \pi 2 i} + e^{\frac \pi{\sqrt 3}-\frac \pi 2 i} + e^{-\frac \pi{\sqrt 3}-\frac \pi 2 i} + e^{\frac \pi{\sqrt 3}+\frac \pi 2 i} \right) \\ & = \frac 16 \left(e^{\frac \pi{\sqrt 3}} + e^{-\frac \pi{\sqrt 3}} + ie^{-\frac \pi{\sqrt 3}} - ie^{\frac \pi{\sqrt 3}} - ie^{-\frac \pi{\sqrt 3}} + ie^{\frac \pi{\sqrt 3}} \right) \\ & = \frac {e^{\frac \pi{\sqrt 3}} + e^{-\frac \pi{\sqrt 3}}}6 \end{aligned}$

Therefore $a = \boxed 3$ .

Σk=1→n π^6(k-1)/27^k=√3^6 I don't know anything about it, but I made it.