Calculate the imit of $\sum_{k=0}^{∞} (-1)^k\frac{1}{k!} $ 

The Taylor series expansion of the function $e^x$ is given by 
\[e^{x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}                  (1)\]
which converges for all $x \in \mathbb{R}$ (the radius of convergence is $\infty$).

Substituting $x=-1$ in (1) we get 
\[\frac{1}{e}=e^{-1}=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}.                  \]