How does $\sum\limits_{n=1}^{\infty} u_n$ converge? 
If $$u_n = \int\limits_{1}^{n} e^{-t^2}\ dt,\ \ \ \ n=1,2,3, \cdots$$then which one of the following statement is TRUE?
$(\text {A})$ Both the sequence $\{u_n \}_{n=1}^{\infty}$ and the series$\sum\limits_{n=1}^{\infty} u_n$ are convergent.
$(\text {B})$ Both the sequence $\{u_n \}_{n=1}^{\infty}$ and the series$\sum\limits_{n=1}^{\infty} u_n$ are divergent.
$(\text {C})$ The sequence $\{u_n \}_{n=1}^{\infty}$ is convergent but the series $\sum\limits_{n=1}^{\infty} u_n$ is divergent.
$(\text {D} )$ $\lim\limits_{n \to \infty} u_n = \frac 2 e.$

This question appeared in GATE exam in the year $2019.$ I found that $$u_n = \sqrt {\pi\left (\frac {1} {e^2} - \frac {1} {e^{2n^2}} \right )},\ \ \ \ n = 1, 2, 3, \cdots$$ Therefore $\lim\limits_{n \to \infty} u_n = \frac {\sqrt {\pi}} {e} \neq 0.$ So the sequence $\{u_n \}_{n=1}^{\infty}$ is convergent but $\sum\limits_{n=1}^{\infty} u_n$ is divergent. So according to me $(\text {C})$ is the correct option although in the answer key it was given that $(\text {A} )$ is the correct option. Am I doing any mistake? Any suggestion regarding this will be highly appreciated.
Thank you very much for your valuable time.
 A: There is no  mistake. C) is the correct answer. $\sum u_n$ is not convergent because $u_n$ does not tend to $0$. 
A: remark  incorrect evaluation of the integral.
Define
$$
u_n = \int\limits_{1}^{n} e^{-t^2}\ dt,\ \ \ \ n=1,2,3, \cdots
$$
I claim
$$
u_n \ne \sqrt {\pi\left (\frac {1} {e^2} - \frac {1} {e^{2n^2}} \right )},\ \ \ \ n = 1, 2, 3, \cdots
$$
It is an error to write
$$
\int_{1}^{n} \int_{1}^{n} e^{-(s^2+t^2)}\ ds\ dt = \int_{\theta = 0}^{2 \pi} \int_{r = \sqrt 2}^{\sqrt 2 n} r e^{-r^2}\ dr\ d\theta.
$$
The left side is the integral over a square.  The right side (in polar coordinates) is not.  
In fact
$$
u_n = \frac{\sqrt{\pi}}{2}\big(\mathrm{erf}(n)-\mathrm{erf}(1)\big)
$$
Numerially
$$
u_2 = \frac{\sqrt{\pi}}{2}\big(\mathrm{erf}(2)-\mathrm{erf}(1)\big)
\approx 0.1352572580
\\
\text{but}\quad\sqrt {\pi\left (\frac {1} {e^2} - \frac {1} {e^{8}} \right )} \approx
0.6512406964
$$

Note that this does not change the fact that $u_n$ does not converge to zero sp that (C) is the right answer.  But of course we can see that without evaluating the integral.
