Showing $e^{-x^2}$ small when $x$ is large just from series definition This is a sort of open ended question. Suppose
$$F(x):=\sum_{L=0}^{\infty} (-1)^L\frac{x^{2L}}{L!}.$$
One can show that the series is absolutely convergent for every fixed $x$, by noting that $x^{2L} \le (L-2)!$ for all large enough $L$. So, $F(x)$ is well defined. I want to show $F(x)$ is very small for $x$ large enough.
I DON'T want to use the fact that $F(x)$ is actually $e^{-x^2}$. Actually, in my research, I am dealing with a similar problem where I have an alternating series quite similar to exponential (but no closed form), where I need to show its small when $x$ is large. So, I was trying with this toy problem.
 A: Here is a method that might work even if you don't have the function in closed form.
First it is seen that since $$F(x):=\sum_{k=0}^{\infty}{\frac{(-1)^kx^{2k}}{k!}}$$
Is a power series with infinite radius of convergence, it has derivatives of all orders that can be obtained by differentiating the terms of the series. In particular,$$F'(x)=\sum_{k=1}^{\infty}{\frac{(-1)^kx^{2k-1}2k}{k!}}=2\sum_{k=1}^{\infty}{\frac{(-1)^kx^{2k-1}}{(k-1)!}}=-2xF(x)$$
for all $x\in \mathbb{R}$. Which can be seen with a shift of index.
$$\\$$
Next we want to show that $1\geq F(x)>0$ on $[0,\infty)$
$F(0)=1>0$. Suppose that $F(x_0)<0$ for some $x_0>0$. Since $F$ is continuous, there is some $c\in(0,x_0)$ such that: $F(c)=0$. We can also assume that there is a largest such $c$, because otherwise $c$ could get arbitrarily close to $x_0$, which would imply that $F(x_0)=0$ via the continuity of $F$.
Since $F<0$ on $(c,x_0)$, then by the relation between $F$ and its derivative, we have that $F'>0$ on $(c,x_0)$. Thus $F$ is increasing on $(c,x_0)$ which implies that $F(x)>F(c)=0$ on $(c,x_0)$. Which is a contradiction.
It can also be shown that $F(x)\neq 0$ on $[0,\infty)$ - I'll leave this to you -
Thus, $0<F(x)\leq 1$ on $[0,\infty)$.
$$\\$$
Lastly,
$$F''(x)=-2xF'(x)-2F(x)=F(x)\left(4x^2-2\right)$$
$$$$
So $x_1=\frac{1}{\sqrt{2}}$ is the only point on $[0,\infty)$ where $F''=0$.
If $0\leq x \leq \frac{1}{\sqrt{2}}\implies F''\leq 0 \implies F'(\frac{1}{\sqrt{2}})\leq F'(x)\leq 0$
If $x\geq \frac{1}{\sqrt{2}}\implies F''\geq 0\implies F'(\frac{1}{\sqrt{2}})\leq F'(x)\leq 0$
Thus $|F'|$ is bounded on $[0,\infty)$.
$$$$
Since, $|F(x)|=\left|\frac{F'(x)}{2x}\right|$, we can now see that $F$ is small if $x$ is large.
A: I like to do unnecessarily complex manipulations on power series for fun, so I thought about this occasionally over the last couple of weeks.  Below I discuss two different solution paths.  The first one is basically done but is unsatisfactory to me.  The second is just some thoughts about an approach that I would find satisfactory.  I'm only including the second part in the hopes of inspiring the OP or someone else to try this approach, since it is nontrivial and (due to the pandemic) there is essentially zero chance that I will have the time to figure it out anytime soon.
Below I'll use $e^{-x}$ instead of $e^{-x^2}$ for simplicity.  I define
$$ f(x) = \sum_n \frac{(-1)^n x^n}{n!} $$
and want to show that $f(x)$ is small when $x$ is large and positive, but without using the fact that $f(x)=e^{-x}$.

Approach 1: This is different than the other answer, and provides a better bound, but has very little chance of working for a different power series.
It's easy to show straight from the power series that $f(x + y) = f(x) f(y)$.  Below, I introduce the index $k=n-m$ to substitute out $n$, at which point the two sums become independent.  (I'm implicitly assuming that we can swap summation order, which I'm not proving, but it's okay because of the convergence properties of the sum.)
$$ \begin{align}
f(x + y) &= \sum_n \frac{(-1)^n (x + y)^n}{n!} \\
&= \sum_n \frac{(-1)^n}{n!} \sum_{m=0}^{n} \binom{n}{m} x^m y^{n-m} \\
&= \sum_n (-1)^n \sum_{m=0}^{n} \frac{x^m y^{n-m}}{m!(n-m)!} \\
&= \sum_m \sum_k \frac{(-1)^{k+m} x^m y^k}{k! m!} \\
&= \left( \sum_m \frac{(-1)^m x^m}{m!} \right) \left( \sum_k \frac{(-1)^k y^k}{k!} \right) \\
&= f(x) f(y)
\end{align} $$
From this, it is obvious that if $n$ is a nonnegative integer, then $f(n) = f(1)^n$.
Since the terms in the series for $f(1)$ are decreasing in magnitude, then we can show that
$$ \frac{1}{3} < f(1) < \frac{1}{2} $$
by just terminating after the third or fourth term in the sum.
Combining these two results, we get that
$$ \frac{1}{3^n} < f(n) < \frac{1}{2^n} $$
for positive integers $n$.
At this point, we want to extend these results to non-integer arguments.  I won't go into details, and there are probably multiple ways of doing this, but one of them would just be to use the same bounds for the derivative, or to place a separate bound on $f(x) - f(n)$ for $x \in (n, n+1)$ using similar means.

As I said, the approach above works, and it gives a reasonably good bound on $f(x)$, but it is extremely unlikely to extend to other series, even approximately.  I feel like it should be possible to show approximate cancellation from the series directly, using some approximations.  I would consider this a "good" solution.
One argument in favor of this, which might be formalizable, is the following.  Consider all of the positive and negative terms in $f(x)$ separately (i.e., the series for $\cosh x$ and $\sinh x$)  The positive ones are $x^{2n}/(2n)!$.  Other than a constant factor, this is the probability of finding value $2n$ in a Poisson distribution with mean $x$; the sum is the probability of finding any even value.  The odd terms are similarly the probability of finding an odd value in the same Poisson distribution.  If the mean is large, then you should be approximately equally likely to find an odd or even value in the Poisson distribution, so basically there should be a lot of cancellation and a small result.
(One way of showing this might be to view the odd and even terms as two different Riemann sums for the integral of the continuous version of the distribution.  But since the distribution is sharply peaked and contains important contributions from many orders of $x$, then it's not obvious to quantify just how small the result should be.)
As another very minor argument in favor of this, it's quite tantalizing that when $x$ is an integer, the two largest-magnitude terms in the series, where $n=x$ and $n=x-1$, cancel exactly.  It's hard to extend this.
I think a general approach, which I have thought about but not tried numerically, is to split the $n^\text{th}$ term in the series into a number of other terms, and rearrange terms in such a way that the result cancels.  (We can do this without fear because the series converges absolutely.)  One way to write this is to say $s_n(x)=(-1)^n x^n/n!$ and write
$$ f(x) = \sum_{n=0}^N s_n(x) + R_N(x) $$
Then we can write the sum as a matrix multiplication,
$$ f(x) = 1^T A s(x) + R_N(x) $$
where $A$ has two properties:

*

*$\sum_j A_{ij} s_j(x) \approx 0$

*$\sum_i A_{ij} = 1$
The first property is basically saying that we weight the different terms in the sum in such a way that each weighted "subseries" cancels exactly, and the second ensures that we get the original sum back when we group the terms back together.  Then the bound on the whole sum would be given by $R_N(x)$ and how good the approximation is in part (1).
There are a number of identities involving the binomial coefficients and/or Stirling numbers that might be useful in showing cancellation of each subseries.  The fact that the exponential is its own derivative suggests that considering finite differences of various orders might be helpful.  I wasn't able to get anything to work, but I'm hoping someone else will.
