Do the roots of $$P_{2n+1}(x)=1+\frac x{1!}+\frac{x^2}{2!}+\cdots+\frac{x^{2n+1}}{(2n+1)!}$$ decrease to $-\infty$?
Can we show this? Indeed, $P_{2n+1}(0)=1$, $P_{2n+1}(-(2n+1))<0$. And $P_{2n+1}'(x)=P_{2n}(x)>0$. So $P_{2n+1}$ has only one root $x_n$. Can we show that $x_n\to-\infty$?
By using, for instance, the Weierstrass $M$-test, we see that your sequence converges uniformly to $e^x$ on any bounded domain. In other words, for any $X< 0$, there is an $N\in \Bbb N$ such that $$|P_{2n+1}(x) - e^x|<\frac{1}{2}e^X \leq \frac12e^x$$ for any $n> N, x\in [X, -X]$. By the triangle inequality, this implies that $P_{2n+1}$ doesn't have any roots on $[X, -X]$ for such $n$. So yes, $x_n\to -\infty$, as for any finite bound we can pick an $N$ which forces $x_n$ to be below that bound whenever $n>N$.
The Weierstrass $M$-test is a very fancy name for a quite simple idea. Basically, say you have a collection of functions $f_n$ for $n\in \Bbb N$ on some domain, and on that domain the absolute value $|f_n|$ of each of those functions is bounded by some number $M_n$. If $\sum M_n < \infty$, then $\sum f_n$ converges uniformly.
In this case, we have $f_n(x) = \frac{x^n}{n!}$, which on $[X, -X]$ is bounded by $M_n = \frac{(-X)^n}{n!}$. Then $\sum M_n = e^{-X}<\infty$, so $\sum f_n$ converges uniformly.