The integral converges to $\sqrt{\pi}/2$. Indeed, let
$$ F(x) = \int_{0}^{x} (1- e^{-1/t^2}) \, dt. $$
By the Abel's test, the series
$$ \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) $$
converges uniformly on $[0,\infty)$ (with the convention $e^{-\infty} = 0$). So we can switch the integration and summation in the following computation:
\begin{align*}
I_R
&:= \int_{0}^{R} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx \\
&\hspace{3em}= \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} \int_{0}^{R} (1 - e^{-(2m+1)^2/x^2}) \, dx\\
&\hspace{6em}= \sum_{m=0}^{\infty} (-1)^m F\left(\frac{R}{2m+1}\right).
\end{align*}
Proceeding,
\begin{align*}
I_R
&= \sum_{m=0}^{\infty} \left\{ F\left(\frac{R}{4m+1}\right) - F\left(\frac{R}{4m+3}\right) \right\} \\
&\hspace{3em}= \sum_{m=0}^{\infty} \int_{\frac{R}{4m+3}}^{\frac{R}{4m+1}} (1 - e^{-1/t^2}) \, dt \\
&\hspace{6em}= \sum_{m=0}^{\infty} \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx,
\end{align*}
where we applied the substitution $x = 1/t$ in the last line. (This substitution is not essential for our argument, but I adopted this step to make clear how monotonicity works.)
Now using the fact that the integrand is decreasing, we can bound $2I_R$ from below by
$$ 2I_R
\geq \sum_{m=0}^{\infty} \left( \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{4m+3}{R}}^{\frac{4m+5}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx \right)
= \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Similar idea shows that
$$ 2I_R \leq \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{1}{R}}^{\frac{3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Finally, taking $R \to \infty$ proves
$$ \int_{0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx
= \lim_{R\to\infty} I_R = \frac{1}{2} \int_{0}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx = \frac{\sqrt{\pi}}{2}. $$
Unless you're forced to use power series, I don't think that power series are the best way to solve this problem. However, if you really want to use power series, note that we can begin by simplifying the integrand using the laws of logarithms to get
$$
\int_0^1\ln\frac{1}{1-x}dx=-\int_0^1\ln(1-x)dx.
$$
Next, observe that this integral is an improper integral since $\ln(1-1)=\ln(0)$ which is undefined. Therefore, we rewrite this as
$$
-\lim_{r\rightarrow 1^-}\int_0^r\ln(1-x)dx.
$$
Note that $\ln(1-x)$ has Taylor series expansion $-\sum_{n=1}^\infty\frac{x^n}{n}$ centered at $0$ with radius of convergence $1$. Therefore, for any $r$ considered above, we can substitute the Taylor expansion to get
$$
\lim_{r\rightarrow 1^-}\int_0^r\sum_{n=1}^\infty\frac{x^n}{n}dx.
$$
Since Taylor (power) series are uniformly convergent within their interval of convergence (this can be made more precise or ignored, depending on your level), we can interchange the limit and the derivative to get that this equals
$$
\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty\int_0^r\frac{x^n}{n}dx=\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty\left.\frac{x^{n+1}}{n(n+1)}\right|_0^r.
$$
Once again, using uniform convergence, we can exchange the sum and the limit to get
$$
\sum_{n=1}^\infty\lim_{r\rightarrow 1^-}\left.\frac{x^{n+1}}{n(n+1)}\right|_0^r=\sum\lim_{r\rightarrow 1^-}\frac{r^{n+1}}{n(n+1)}=\sum_{n=1}^\infty\frac{1}{n(n+1)}.
$$
Finally, we can use a partial fraction decomposition to write this as
$$
\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right),
$$
which telescopes to $1$.
Now, if I were solving this without any restrictions, I would take the integral
$$
-\int_0^1\ln(1-x)dx
$$
and use the $u$-substitution $u=1-x$ so $du=-dx$ to get
$$
\int_1^0\ln(u)du.
$$
This integral is, once again, improper when $u=0$, so we write
$$
\lim_{r\rightarrow 0^+}\int_1^r\ln(u)du.
$$
An antiderivative of $\ln(u)$ is $u\ln(u)-u$, which can be found through integration by parts, for example. Therefore, this simplifies to
$$
\lim_{r\rightarrow 0^+}(u\ln u-u)|_1^r=\lim_{r\rightarrow 0^+}(r\ln r-r)-(1\ln 1-1).
$$
Now, $\lim_{r\rightarrow 0^+}r\ln r$ is an indeterminate form, but an application of l'Hopital's rule results in a value of $0$, so this entire expression simplifies to $1$, as expected.
Best Answer
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
$$f_n(x) = \begin{cases} (1 - \frac{x^2}{n})^n, & x \le \sqrt{n}\\0, & x \ge \sqrt{n}\end{cases}$$
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 $x < \sqrt{n}$, 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:
$$\begin{align}\frac{\sqrt{\pi}}{2} = \int_0^\infty e^{-x^2} dx &= \int_0^\infty \lim_{n\to\infty} f_n(x) dx \stackrel{\rm MCT}{=} \lim_{n\to\infty} \int_0^\infty f_n(x) dx\\ &= \lim_{n\to\infty} \int_0^{\sqrt{n}}\left(1 - \frac{x^2}{n}\right)^n dx = \lim_{n\to\infty} \sqrt{n} \int_0^1 \left(1 - t^2\right)^n dt\\ &= \lim_{n\to\infty} \sqrt{n}\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k} \end{align}$$