Gaussian integral via power series? It is well known that the Gaussian integral is $\sqrt \pi$, i.e.,
$$ \int_0^\infty e^{-x^2} dx = \frac{\sqrt \pi}2.$$
This is something we typically learn in calculus with polar coordinates.  We also learn how to integrate $e^{-x^2}$ via power series, e.g.,
$$ \int_0^t e^{-x^2} dx = \sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}.$$
Hence one has 
$$ \frac{\sqrt \pi}2 = \lim_{t \to \infty} \left(\sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}\right).$$

Question: Is there a nicer way to rewrite the expression on the right, e.g. with "only one infinity" involved, to get an interesting expression/algorithm for computing $\pi$?

Of course the goal would be "pass the limit through the sum" but this is evidently non-trivial as the tail of the series explodes with $t$.  My limited experience with analysis suggests one could try to split up the sum in various ranges ($n \ll t$, $n \approx t$, $n \gg t$) but that middle ranges could get hairy.
In addition to changing the "upper bound" $t$ of integral,
you can make the integrand depends on $t$. This allows you
to get rid of the tail of the integrand for finite $t$.

For any $n \in \mathbb{Z}_{+}$, let $f_n : [0,\infty) \to \mathbb{R}$ be the function

$$f_n(x) = \begin{cases} (1 - \frac{x^2}{n})^n, &amp; x \le \sqrt{n}\\0, &amp; x \ge \sqrt{n}\end{cases}$$

For any fixed $x \in [0,\infty)$, it is easy to see $f_n(x) \le f_{n+1}(x)$ whenever $x \ge \sqrt{n}$.
When $x &lt; \sqrt{n}$, we can apply AM $\ge$ GM to $n$ copies of $1 - \frac{x^2}{n}$ and one copy of $1$ and get

$$1 - \frac{x^2}{n+1} = \frac{n}{n+1}\left(1 - \frac{x^2}{n}\right) + \frac{1}{n+1} \ge \left( 1 - \frac{x^2}{n}\right)^{n/n+1} \implies f_{n+1}(x) \ge f_n(x)$$

This means $f_1(x), f_2(x), \ldots$ is a sequence of pointwise non-decreasing,
non-negative functions. Since its pointwise limit equals to $e^{-x^2}$, we can use Monotone convergence theorm to convert the integral on $e^{-x^2}$ to a limit of single variable:

$$\begin{align}\frac{\sqrt{\pi}}{2} 
= \int_0^\infty e^{-x^2} dx 
&amp;= \int_0^\infty \lim_{n\to\infty} f_n(x) dx
\stackrel{\rm MCT}{=}
\lim_{n\to\infty} \int_0^\infty f_n(x) dx\\
&amp;= \lim_{n\to\infty} \int_0^{\sqrt{n}}\left(1 - \frac{x^2}{n}\right)^n dx
= \lim_{n\to\infty} \sqrt{n} \int_0^1 \left(1 - t^2\right)^n dt\\
&amp;= \lim_{n\to\infty} \sqrt{n}\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}
\end{align}$$
The Expression Can Be Written As Zeta Function and Zeta Function At infinity is 1 hence using proper methods it may be evaluated