Summation of$ n^3\over (3n)!$

Question

Summation of $n^3\over(3n)!$ from 1 to infinity

Please provide procedure.

Don't know how to proceed.

Tried making it to form $n^2\over3 (3n-1)!$ then?

Answer 1

Let $\zeta_3$ denote a third root of unity. Consider the sum \begin{equation*} e^x + e^{\zeta_3 x} + e^{\zeta_3^2 x} = \sum_{k = 0}^{\infty} \frac{x^k}{k!}(1+\zeta_3^k + \zeta_3^{2k}). \end{equation*} The bracketed term is non-zero if and only if $3|k$ (say by a geometric series argument or just from orthogonality of characters or the like). Note further that if $f(x) := \sum_{k = 0}^{\infty} a_kx^k$ is a power series then by formal differentiation and multiplication $xf'(x) = \sum_{k \geq 0} ka_kx^k$, and repeating this twice more yields $\sum_{k \geq 0} k^3a_kx^k$. Therefore, differentiate then multiply by $x$ the sum of exponentials above and you should get on one hand the sum $\sum_{k \geq 0} \frac{x^{3k}(3k)^3}{(3k)!}$ and then plug in $x = 1$.

Answer 2

HINT:

$$\frac{n^3}{(3n)!}=\frac1{27}\frac{(3n)^3}{(3n)!}$$

Now set $x=1,\omega,\omega^2$ in $\displaystyle e^x=\sum_{r=0}^\infty\frac{x^r}{r!}$ and add where $\omega^3=1$ and $1+\omega+\omega^2=0$

We can utilize Euler Formula to find $e^\omega, e^{\omega^2}$

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Alternatively, $$\frac{n^3}{(3n)!}=\frac A{(3n)!}+\frac B{(3n-1)!}+\frac C{(3n-2)!}+\frac D{(3n-3)!}$$

$$\implies n^3=A+B(3n)+C 3n(3n-1)+D3n(3n-1)(3n-2)$$

Setting $3n=0\iff n=0$ in the above identity, $0=A$

$\displaystyle3n-1=0\iff n=\frac13,\left(\frac13\right)^3=3B\cdot\left(\frac13\right)\iff B=\frac1{27}$

Similarly find $C,D$

Now utilize $\displaystyle e^x=\sum_{r=0}^\infty\frac{x^r}{r!}$