Why does $1 - x + \frac{1}{2}x^2-\frac{1}{6}x^3+\dots$ converge to $e^{-x}$? The series
$$
1 - x + \frac{1}{2}x^2-\frac{1}{6}x^3+\dots=e^{-x}
$$
It is not totally clear from (for example) limits like
$$
\lim_{x \to \infty}( 1 - x + \frac{1}{2}x^2) = \infty \\
\lim_{x \to \infty}( 1 - x + \frac{1}{2}x^2-\frac{1}{6}x^3) = -\infty \\
\vdots
$$
why the sum to infinity has a limit of 0. What is "happening" after each partial sum diverges? I suppose the remaining part of the series diverges in the other direction in just the right way that we end up with facts like
$$
\lim_{x \to \infty}( e^{-x}) = 0
$$
but this is not always true just from the oscillating infinities.
The reason this is important is that there should/might be a general condition for a oscillating power series to have a limit of $0$, of which this is a simple example. But how does even this simple case have limit $0$, without first seeing that the series is $e^{-x}$?
 A: I do not think there is a simple procedure for establishing if
$$ \lim_{x\to +\infty}\sum_{n\geq 0}(-1)^n a_n x^n $$
equals zero or not, provided that $\{a_n\}_{n\geq 0}$ is a non-negative sequence convergent to zero rapidly enough to make the function entire. For instance, in the case $a_n = \frac{1}{n!^2}$ we have
$$ \sum_{n\geq 0}(-1)^n a_n x^n = O\left(\frac{1}{\sqrt[4]{x}}\right)\quad\text{as }x\to +\infty\tag{1} $$
which looks pretty non-trivial to me. Such things (like Tricomi's inequality hidden in $(1)$) are usually proved by exploiting the fact that the structure of the sequence $\{a_n\}_{n\geq 0}$ makes the generating function a solution of a reasonably simple differential equation. For instance, by defining
$$ \sinh(x) = \sum_{n\geq 0}\frac{x^{2n+1}}{(2n+1)!} $$
$$ \cosh(x) = \sum_{n\geq 0}\frac{x^{2n}}{(2n)!} $$
we have
$$ \lim_{x\to +\infty}\sum_{n\geq 0}\frac{(-1)^n}{n!}x^n = \lim_{x\to +\infty}\left(\cosh(x)-\sinh(x)\right) $$
and the structure of $a_n=\frac{1}{n!}$ gives the nice identity $\cosh^2(x)=1+\sinh^2(x)$ reminiscent of the Pythagorean theorem. That also gives that the previous limit equals
$$ \lim_{x\to +\infty}\left(\sqrt{1+\sinh^2(x)}-\sinh(x)\right)=\lim_{x\to +\infty}\frac{1}{\sinh(x)+\cosh(x)} $$
and now this is blatantly zero since both $\sinh(x)$ and $\cosh(x)$ are increasing and unbounded over $\mathbb{R}^+$.
A: First of all, the series
$$\sum_{n=0}^\infty (-1)^{^n}\dfrac{x^n}{n!}$$
converges absolutely in $\mathbb{R}$, by the ratio test, so we can denote
$$f(x)=\sum_{n=0}^\infty (-1)^{n}\dfrac{x^n}{n!}$$
Now, absolute convergence allows as to conclude that the Cauchy product of $f(x)$ and $f(y)$ converges to $f(x)f(y)$, and their Cauchy product is
$$\sum_{n=0}^\infty \left( \sum_{k=0}^n \left((-1)^{k}\dfrac{x^k}{k!}\right)\left( (-1)^{n-k}\dfrac{y^{n-k}}{(n-k)!} \right) \right) = \sum_{n=0}^\infty \dfrac{(-1)^n}{n!} \sum_{k=0}^n \binom{n}{k} x^n y^{n-k} = \sum_{n=0}^\infty (-1)^n \dfrac{(x+y)^n}{n!} = f(x+y)$$
So, we have prove that $f(x+y)=f(x)f(y)$ and letting $y=-x$ we get that
$$f(x)f(-x)=f(0)=1$$
and so
$$f(-x)=\dfrac{1}{f(x)} \forall x \in \mathbb{R}$$
Then
$$\lim_{x \to +\infty} f(x) = \lim_{x \to -\infty} f(-x)= \lim_{x \to -\infty} \dfrac{1}{f(x)}$$
and is enough to see that
$$\lim_{x \to -\infty} f(x) = +\infty$$
But, for any $x<0$ we have that
$$f(x)=f(-|x|) = \sum_{n=0}^\infty (-1)^n \dfrac{(-|x|)^n}{n!} = \sum_{n=0}^\infty \dfrac{|x|^n}{n!} \geq 1+|x|$$
and as
$$\lim_{x \to -\infty} 1+ |x|=+\infty$$
we get the desired result.
A: Let's call $P_n(x)$ the polynomial function made of partial sums
$$
P_0(x) = 1
$$
$$
P_1(x) = 1-x
$$
$$
P_2(x) = 1-x+\dfrac{1}{2}x^2
$$
$$
\vdots
$$
$$
P_{n}(x) = \sum_{i=0}^{n} \dfrac{(-1)^i}{i!} \cdot x^{i}
$$
The infinite series in a domain $D$ using the partial polynomials normally is defined by the limit
$$
P(x) = \lim_{n \to \infty} P_{n}(x) \ \ \ \forall x \in D
$$
And when you do a limit $x \to \infty$ of $P(x)$ you are computing
$$
\lim_{x \to \infty} P(x) = \lim_{x \to \infty} \lim_{n \to \infty} P_{n}(x) \overset{?}{=} \lim_{x \to \infty \\ n \to \infty} P_{n}(x) \overset{?}{=} \lim_{n \to \infty} \lim_{x \to \infty} P_{n}(x) 
$$
In this example you are taking the limit of two variables and rearranging they as you want. Depending on your choice of path, the limit of two variables doesn't converge and the 'notion' of series is not the same anymore.
A: I think that the proof is simpler than it seems. We write
$1-x+\dfrac{1}{2}x^{2}-\dfrac{1}{3!}x^{3}+...+\dfrac{1}{n!}x^{n}+\sum_{k=n+1}^{+\infty}(-1)^{k}\dfrac{1}{k!}x^{k}-e^{-x}$
and now we use Taylor's theorem with a remainder for $e^{-x}$=$1-x+\dfrac{1}{2}x^{2}-\dfrac{1}{3!}x^{3}+...+(-1)^{n}\dfrac{1}{n!}x^{n}$$+(+/-)$$\dfrac{e^{-\xi}}{(n+1)!}$$x^{n+1}$. After subtracting we are left with
$\sum_{k=n+1}^{+\infty}(-1)^{k}\dfrac{1}{k!}x^{k}$+$(+/-)\dfrac{e^{-\xi}}{(n+1)!}$$x^{n+1}$. It is well known that $\dfrac{x^{n+1}}{(n+1)!}$ $\to 0$ and clearly
$\sum_{k=n+1}^{+\infty}(-1)^{k}\dfrac{1}{k!}x^{k}$ $\to 0$ since the series is convergent.
So on any closed bounded set $B$ such that $x\in B$ we have uniform convergence, and the result is proved!!
A: The answer of your question "is alternating  between $\infty$ and $-\infty$ is enough to say that the limit of the series will be $0$"? is NO.
It works for the $e^{-x}$ series. But it does not work for $\sin (x)$ for example.
