prove $\int_0^{\infty} e^{-t^2}dt $ converges Show the convergence of :  $$\int_0^{\infty} e^{-t^2}dt $$
I believe I can prove this with the Cauchy integral convergence test but was looking for any simpler method.
 A: It suffices to observe that on $[0,1)$, $$f(t) = e^{-t^2} \le e^0 = 1,$$ and on $[1,\infty)$, $$f(t) \le te^{-t^2},$$ consequently $$\int_{t=0}^\infty f(t) \, dt \le \int_{t=0}^1 dt + \int_{t=1}^\infty te^{-t^2} \, dt = 1 + \lim_{n \to \infty} \left[-\frac{e^{-t^2}}{2} \right]_{t=1}^n = 1 + \frac{1}{2e}.$$
A: Note that
$$\sum_{n=0}^\infty e^{-n}=\frac{1}{1-e^{-1}},$$
which shows that $\displaystyle\sum_{n=0}^\infty e^{-n}$ converges. Also, it is easy to see that $e^{-n^2}\le e^{-n}$ for each $n\in\mathbb{N}\cup\{0\}$, thus
$$\sum_{n=0}^\infty e^{-n^2}\le\sum_{n=0}^\infty e^{-n}$$
and by using the comparison test, $\displaystyle\sum_{n=0}^\infty e^{-n^2}$ converges, and so is $\displaystyle\int_0^\infty e^{-t^2}dt$.
A: This can be done in a rather easy way. On interval $(0,1)$ the given function is convergent and on interval $1$ to infinity it is easy to see that your function is less than the function  $y=e^{-t}$ The latter integral is very easy to integrate on interval $1$ to infinity. In fact that value is $1/e$. 
