Problem: If
\[
u_n = \int_1^n e^{-t^2}\mathrm{d} t,\quad n = 1,2,3,\dots,
\]
then which one of the following statements is TRUE?
-
Both the sequence $\{ u_n \} _{n=1}^{\infty} $ and the series $\displaystyle \sum_{n=1}^{\infty} u_n$ are convergent.
-
Both the sequence $\{ u_n \} _{n=1}^{\infty} $ and the series $\displaystyle \sum_{n=1}^{\infty} u_n$ are divergent.
-
The sequence $\{ u_n \} _{n=1}^{\infty} $ is convergent but the series $\displaystyle \sum_{n=1}^{\infty} u_n$ is divergent.
-
$\displaystyle \lim_{n \to \infty} u_n = \frac{2}{e}$.
Solution: The given sequence is
\[
u_n = \int_1^n e^{-t^2}\mathrm{d} t.
\]
The above integral can be written as
\begin{align*}
u_n & = \int_1^n e^{-t^2}\mathrm{d} t + \int_0^1 e^{-t^2}\mathrm{d} t - \int_0^1 e^{-t^2}\mathrm{d} t \\[1ex]
& = \int_0^n e^{-t^2}\mathrm{d} t - \int _0^1 e^{-t^2}\mathrm{d} t \\[1ex]
\implies \lim_{n \to \infty} u_n & = \lim_{n \to \infty} \int_0^n e^{-t^2}\mathrm{d} t - \int_0^1 e^{-t^2}\mathrm{d} t \\[1ex]
& = \int_0^{\infty}e^{-t^2}\mathrm{d} t - \int_0^1 e^{-t^2}\mathrm{d} t \\[1ex]
& = \frac{\sqrt{\pi } }{2} - k,
\end{align*}
where $k = \int_0^1 e^{-t^2}\mathrm{d} t$. Therefore, the sequence $\{ u_n \} $ is convergent.
It is clear that $k \neq \frac{\sqrt{\pi } }{2}$, so the sequence $\{ u_n \} $ does not converges to $0$ and hence the series $\sum_{u_n} $ is divergent. Therefore, the correct answer will be OPTION C.