Let $f(t) = e^{-|t|}$, $t \in \mathbb{R}$. The function $f$ has the everywhere convergent power series expansion
$$f(t) = \sum_{n=0}^\infty \frac{(-1)^n|t|^n}{n!}.$$
I would like to know whether there exists some $C > 0$ independent of $t$ so that for all integers $M \ge 0$, $$\Big|\sum_{n=0}^M \frac{(-1)^n|t|^n}{n!} \Big| \le C.$$
If we were to instead consider the power series expansion, for, say, $e^t$, then we do not expect there to be a uniform constant, because $e^t$ becomes too large as $t \to \infty$. However, since $f$ is decaying at both $\pm \infty$, I suspect such a $C$ can be found, and one will need to rely somehow on the cancellation of the positive and negative terms in the series.
Hints or solutions are greatly appreciated!
For any $M>0$ the term $\sum_{n=0}^M \frac{(-1)^n|t|^n}{n!}$ is a polynomial in $|t|$ which is unbounded as $|t| \to \infty$.