If $u_n= \int_1^ne^{-t^2}dt$, what can be said about $\sum u_n$? If

$\displaystyle u_n= \int_1^ne^{-t^2}dt$ where $n=1,2,3...$

Then Which of the following is true

$1)$ both the sequence $u_n$ and series $\sum u_n$ is convergent
$2)$ both the sequence $u_n$ and series $\sum u_n$ is divergent
$3)$ The sequence is convergent and $\sum u_n$ is divergent
$4)$ $\displaystyle \lim_{n \to \infty}u_n=\frac{2}{e}$

The solution I tried - I know that $\displaystyle\int_0^{\infty} e^{-t^2}=\sqrt  \pi$ ,but how to calculate the $u_2,u_3...$ I have no idea please help
 A: The sequence $u_n$ is convergent to $C=\int_{1}^{+\infty}e^{-t^2}\,dt$. Because of this, the series $\sum u_n$ is divergent.
No quantitative estimation is really needed, but if you want one
$$ u_n = C-\int_{n}^{+\infty}e^{-t^2}\,dt = C-\frac{1}{2}\int_{n^2}^{+\infty}\frac{dt}{e^t\sqrt{t}}\geq C-\frac{1}{2n}\int_{n^2}^{+\infty}\frac{dt}{e^t}=C-\frac{1}{2ne^{n^2}}. $$
A: The answer is 3).


*

*The sequence is convergent, because we have
$$\lim_{n \rightarrow \infty} \left(\int_{1}^{n} e^{-t^2}\ dt\right) = \int_{1}^{\infty} e^{-t^2} dt = \left(\int_{0}^{\infty} e^{-t^2} dt\right) - \left(\int_{0}^{1} e^{t^2}\ dt\right)$$
and the left hand integral is $\sqrt{\pi}$, while the right-hand one must be a finite value(*) because $t \mapsto e^{-t^2}$ has no singularities for any $t \in [0, 1]$. Hence the limit exists by virtue of this expression thus having a well-defined finite value.

*The series is divergent. This follows from the fact the above limit is nonzero, as it is a basic theorem that a necessary (but not sufficient) condition for convergence of a series is that the sequence of terms involved must converge to zero at infinity. The fact the limit is nonzero, in turn, is because in
$$\int_{1}^{\infty} e^{-t^2}\ dt$$
the integrand $t \mapsto e^{-t^2}$ is always positive in $[1, \infty)$ (and anywhere else, for that matter).



(*) For what it's worth, the exact value is $\frac{\sqrt{\pi}}{2} \left(1 - \mathrm{erf}(1)\right) \approx 0.7468$ (though one might object this is a bit "circular" in a sense, but...).
