Rewrite the $k$th term using $k(k+1)=k(k-1)+2k$:

$T_k=(-1{)}^{k+1}\frac{k(k+1)}{k!}=(-1{)}^{k+1}\frac{k(k-1)+2k}{k!}=\frac{(-1{)}^{k+1}}{(k-2)!}+\frac{2(-1{)}^{k+1}}{(k-1)!}$ (for $k\ge 2$ in the first piece; the $k=1$ term contributes $0$ to it).

$\sum _{k=1}^{\infty }{T}_k=\sum _{k=2}^{\infty }\frac{(-1{)}^{k+1}}{(k-2)!}+2\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{(k-1)!}$

Shift indices ($j=k-2$ in the first, $j=k-1$ in the second):

$={\sum }_{j=0}^{\infty }\frac{(-1{)}^{j+1}}{j!}+2{\sum }_{j=0}^{\infty }\frac{(-1{)}^j}{j!}=-e^{-1}+2e^{-1}=e^{-1}=\frac{1}{e}$.

Option (2).