Happy summing

$\displaystyle \sum^{n-1}_{m=0}3^m =\dfrac{3^n-1}{2}~~(\color{#D61F06}{Sum~of~Infinite~GP})$ Hence summation simplifies to: $\Large\mathfrak{S}=\dfrac{1}{2}\displaystyle \sum^{\infty}_{n=1} (\color{#20A900}{\dfrac{ 3^n-1}{n!}})$ $\Large =\dfrac{1}{2}(e^3-e)\approx 8.68$ $\huge \therefore \lfloor 8.68 \rfloor=\boxed{\color{#007fff}{8}}$

$\large\boxed{\mathcal{\color{#302B94}{NOTE:-}}\\ \color{#3D99F6}{e^x-1=\displaystyle\sum_{n=1}^{\infty}\dfrac{x^n}{n!}}\\ \text{Put x=3 and x=1 and then subtract}\\ \color{#20A900}{\Rightarrow e^3-e=\displaystyle\sum_{n=1}^{\infty}\dfrac{3^n-1}{n!}}}$