Answer
TRUE
Work Step by Step
Since, $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$
Put the value of $x$ into the series as $\frac{1}{e}=e^{-1}$, so we can plug in $x=-1$
Therefore,
$e^{-1}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$
OR
$\frac{1}{e}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$
Hence, the statement is TRUE.
