Why not integrate it !!

$\large\mathfrak{T}=\sum_{k=1}^{n} \frac{(k+3)(k+2)(k+1)-1}{ \left (k+3 \right)!}$

$\large =\sum_{k=1}^{n}\left(\dfrac{1}{k!}-\dfrac{1}{(k+3)!}\right)(**)$

$\mathbf{(A~Telescopic~ Series)}$

$\therefore\large \small{\displaystyle\lim_{n\to \infty}\mathfrak{T}}=\left( \begin{aligned} & ~~~~~\left(1-\cancel{\color{#3D99F6}{\dfrac{1}{4!}}}\right)\\&+\left(\dfrac{1}{2!}-\cancel{\color{#D61F06}{\dfrac{1}{ 5!}}}\right)\\&+\left(\dfrac{1}{3!}-\cancel{\color{#456461}{\dfrac{1}{6!}}}\right) \\&+\left(\cancel{\color{#3D99F6}{\dfrac{1}{4!}}}- \cancel{\color{#EC7300}{\dfrac{1}{7!}}}\right)\\&\cdots\\&\cdots\end{aligned} \right)$

$\Large =1+\dfrac 12+\dfrac{1}{6}=\dfrac 53$

$\huge \therefore 5+3=\boxed{8}$

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$**$ Alternatively we can use expansion of $e$ i.e $e=1+\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n!}$

$\displaystyle\lim_{n\to \infty}\mathfrak{T}=\sum_{k=1}^{\infty}\left(\dfrac{1}{k!}\right)- \sum_{k=4}^{\infty}\left(\dfrac{1}{k!}\right)$

$=(e-1)-(e-1-1-\dfrac{1}{2}-\dfrac{1}{6})=\dfrac{5}{3}$

Thus we see that the limit converges and its nothing but $e !!$ . This was the method in which I solved, I want to know if people have any other solution. Thanks for the solution again