Sum of Series S =? 
Proving:
$$\displaystyle\sum_{n=0}^{\infty}\frac{(n^2+4n-1)(\pi-\phi)^{3n}}{n!(n+2)}=(\pi-\phi)^3e^{(\pi-\phi)^3}+2\left(e^{(\pi-\phi)^3}-1\right)-5\left(\frac{e^{(\pi-\phi)^3}((\pi-\phi)^3-1)+(\pi-\phi)^4-(\pi-\phi)^3+1}{(\pi-\phi)^6}\right) $$
Where .$ \phi $ is  golden ration

Note:this series is convergent
Proof :
we put $$S=\displaystyle\sum_{n=0}^{\infty}\frac{n^2+4n-1}{n!(n+2)}x^{3n}$$
we have :
\begin{align*}
\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|
&=\lim_{n\to\infty}|\frac{\frac{(n+1)^2+4(n+1)-1}{(n+1)!(n+3)}}{\frac{n^2+4n-1}{n!(n+2)}}|\\
&=\lim_{n\to\infty}\frac{(n+1)^2+4(n+1)-1}{n^2+4n-1}\frac{(n+2)}{(n+1)(n+3)}=0\\
&\implies R=\infty
\end{align*}
So This series is convergent for each $x\in \mathrm R$
 A: Note that for $n\geq 1$,
$$
\frac{{n^2  + 4n - 1}}{{n!(n + 2)}}x^{3n}  = \frac{{x^{3n} }}{{(n - 1)!}} + 2\frac{{x^{3n} }}{{n!}} - 5\frac{{x^{3n} }}{{n!(n + 2)}}.
$$
Hence,
$$
\sum\limits_{n = 0}^\infty  {\frac{{n^2  + 4n - 1}}{{n!(n + 2)}}x^{3n} }  = e^{x^3 } (x^3  + 2) - 5\sum\limits_{n = 0}^\infty  {\frac{{x^{3n} }}{{n!(n + 2)}}} .
$$
But
\begin{align*}
\sum\limits_{n = 0}^\infty  {\frac{{x^{3n} }}{{n!(n + 2)}}} & = \frac{1}{{x^6 }}\sum\limits_{n = 0}^\infty  {\frac{{x^{3(n + 2)} }}{{n!(n + 2)}}}  = \frac{1}{{x^6 }}\sum\limits_{n = 0}^\infty  {\frac{{(n + 1)x^{3(n + 2)} }}{{(n + 2)!}}} \\ & = \frac{1}{{x^3 }}\sum\limits_{n = 0}^\infty  {\frac{{x^{3(n + 1)} }}{{(n + 1)!}}}  - \frac{1}{{x^6 }}\sum\limits_{n = 0}^\infty  {\frac{{x^{3(n + 2)} }}{{(n + 2)!}}} \\ & = \frac{1}{{x^3 }}(e^{x^3 }  - 1) - \frac{1}{{x^6 }}(e^{x^3 }  - 1 - x^3 ) = \frac{{e^{x^3 } (x^3  - 1) + 1}}{{x^6 }}.
\end{align*}
Accordingly,
$$
\sum\limits_{n = 0}^\infty  {\frac{{n^2  + 4n - 1}}{{n!(n + 2)}}x^{3n} }  = e^{x^3 } (x^3  + 2) - 5\frac{{e^{x^3 } (x^3  - 1) + 1}}{{x^6 }}.
$$
Note that this expression differs from your result. The term $(\pi-\phi)^4$ in your answer is suspicious to me. Please check it numercially again.
