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.

  • $\begingroup$ If true, that's a pretty egregious mistake in the answer sheet. Are you sure the limits in the integral defining $u_n$ weren't $\int_n^{n+1}$? $\endgroup$ – Clement C. Dec 20 '19 at 6:35
  • $\begingroup$ Yes. I am pretty sure @Clement C. $\endgroup$ – math maniac. Dec 20 '19 at 6:36
  • $\begingroup$ gate.iitd.ac.in/OldQP/2019QP_MA.pdf look at question no. $5$ @Clement C. $\endgroup$ – math maniac. Dec 20 '19 at 6:40
  • $\begingroup$ Well, yes. The right answer is (C). The answer key is wrong. There's not much more to say. $\endgroup$ – Clement C. Dec 20 '19 at 6:51
  • $\begingroup$ Your evaluation of the integral is incorrect. So nothing after that point contributes to an answer. $\int e^{-t^2}\;dt$ is not an elementary function. $\endgroup$ – GEdgar Dec 20 '19 at 11:40

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.

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  • $\begingroup$ Would you please add something more to your answer @GEdgar e.g. what is $\text {erf},$ what is the correct way to evaluate the integral, why transformation to polar coordinates fails here, when do I transform cartesian to polar coordinates etc. Then it will be very helpful for me. Thanks for your valuable suggestion. I am very surprised that nobody except you pointed my mistake out. You are kind enough to make me aware of erroneous step I did in evaluating this integral. $\endgroup$ – math maniac. Dec 20 '19 at 18:00
  • $\begingroup$ Furthermore I will be more than happy if you kindly suggest me to read some books which cover this topic. Thanks again @GEdgar. $\endgroup$ – math maniac. Dec 20 '19 at 18:05
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    $\begingroup$ $\mathrm{erf}$ or "error function" en.wikipedia.org/wiki/Error_function , evaluation of your integral is essentially the definition of $\mathrm{erf}$ . Calculus books cover change of variables, especially using polar coordinates. You did it correctly, except for the upper and lower bounds on the integral signs. Surely $\theta$ does not go all the way around the circle? And $r=\sqrt{2}$ is the lower bound only when $\theta = \pi/4$, not other angles. Draw pictures to see this. $\endgroup$ – GEdgar Dec 20 '19 at 23:55
  • $\begingroup$ You are absolutely right @GEdgar. There is no way we can match up $r$ and $\theta$ to get the required region (i.e. the square of side lengths $n-1$) in which we are trying to evaluate our integral. Whatever $r$ and $\theta$ we are going to take it will be an annulus or portion of an annulus. So we cannot ever keep track of the region of integration we require by using polar coordinates. This trick is only valid when we integrate the entire integral over the entire real line because then the induced double integral has to be taken over the entire $x,y$-plane. $\endgroup$ – math maniac. Dec 21 '19 at 5:49
  • $\begingroup$ And the nice thing is that we can easily keep track of the entire plane by setting the limit of $r$ between $0$ to $\infty$ and that the limit of $\theta$ from $0$ to $2 \pi.$ I understood your point quite clearly now. The bottom line of the story $:$ We can use polar coordinates to evaluate the double integral when we are able to keep track of the region we are interested in by means of $r$ and $\theta.$ $\endgroup$ – math maniac. Dec 21 '19 at 5:58

There is no mistake. C) is the correct answer. $\sum u_n$ is not convergent because $u_n$ does not tend to $0$.

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  • $\begingroup$ Then why in the answer key it was given that $(\text {A})$ is the correct option @KaviRamaMurthy sir? $\endgroup$ – math maniac. Dec 20 '19 at 6:35
  • $\begingroup$ @mathmaniac. Mistakes are made? $\endgroup$ – Clement C. Dec 20 '19 at 6:35
  • $\begingroup$ But for that we all were given negative marks @Clement C. $\endgroup$ – math maniac. Dec 20 '19 at 6:36
  • $\begingroup$ What can I say... that's bad. Try to dispute the grading with the instructor? $\endgroup$ – Clement C. Dec 20 '19 at 6:37
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    $\begingroup$ @mathmaniac. Nothing much can be done about it. Sorry that such things are happening. $\endgroup$ – Kavi Rama Murthy Dec 20 '19 at 7:15

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