Compute $$\lim _{n\to \infty }\left(\frac{\left(4n-4-2a_n\right)}{\pi }\right)^n$$

Where $$a_n=\int _1^n\:\frac{2x^2}{x^2+1}dx$$

The integral I solved and I got $a_n=2(x-\arctan(x))$ Afterwards, after I took into account the fact that it's a definite integral, I got $a_n=2(n-\arctan(n)-\frac{\left(4-\pi\right)}{4})$

I tried to meddle with the limit a bit too and I got $\lim _{n\to \infty }3^n$, which would be $\infty$. Is this correct? Since it seemed way too easy. Could someone check my answer?

  • $\begingroup$ The limit should be $e^{-4/\pi}$ $\endgroup$ – Zain Patel Jul 5 '15 at 14:29
  • $\begingroup$ That doesn't look correct to me offhand, no. Why don't you write out $$ \left(\frac{\left(4n-4-2a_n\right)}{\pi }\right)^n $$ for us by substituting in your value for $a_n$ and simplifying. $\endgroup$ – John Hughes Jul 5 '15 at 14:29
  • $\begingroup$ I don't thing the integral you got is correct. $\endgroup$ – Timbuc Jul 5 '15 at 14:29
  • $\begingroup$ $a_n$ is not a function of $x$, it should be a function of $n$. Don't even write $a_n=2(x-\arctan x)$. $\endgroup$ – Thomas Andrews Jul 5 '15 at 14:29
  • $\begingroup$ So, $2a_n = 4n-4-4\arctan x+\pi$ so $4n-4-2a_n = 4\arctan n -\pi$. $\endgroup$ – Thomas Andrews Jul 5 '15 at 14:32


$$a_n=\left.\int_1^n\frac{2x^2}{1+x^2}dx=\left(2x-2\arctan x\right)\right|_1^n=2n-2\arctan n-2+\frac\pi2\implies$$

$$\frac{4n-4-2a_n}\pi=\frac{4\arctan n}\pi-1\implies$$


For the last limit, you may want to use the exponential function:

$$\lim_{n\to\infty}n\log\left(\frac{4\arctan n}\pi-1\right)=\lim_{n\to\infty}\frac{\log\left(\frac{4\arctan n}\pi-1\right)}{\frac1n}\stackrel{l'H}=$$

$$=\lim_{n\to\infty}\frac{\frac\pi{4\arctan n-\pi}\frac4{\pi(1+n^2)}}{-\frac1{n^2}}=-\frac4\pi$$

  • $\begingroup$ Why spend so much time repeating the first part that the OP has already done? $\endgroup$ – Thomas Andrews Jul 5 '15 at 14:45
  • $\begingroup$ @ThomasAndrews (1) Because I didn't notice, (2) Because after I notice (now), I can't see clearly whether it is the same thing. (3) It is not "so much time"...at least not for me :) $\endgroup$ – Timbuc Jul 5 '15 at 14:47
  • $\begingroup$ But why is that limit at the end $\frac{-4}{\pi }$? :S $\endgroup$ – MikhaelM Jul 5 '15 at 15:10
  • $\begingroup$ @MikhaelM This is simply a limit...not even a hard one: $$\frac{-4}{4\arctan n-\pi}\cdot\frac{n^2}{1+n^2}\xrightarrow[n\to\infty]{}\frac{-4}{4\frac\pi2-\pi} \cdot1=-\frac4\pi$$ $\endgroup$ – Timbuc Jul 5 '15 at 15:17
  • $\begingroup$ Thank you @Timbuc . I still got a lot of work to do with limits :) $\endgroup$ – MikhaelM Jul 5 '15 at 15:23


$$\lim_{n\to \infty} \left(\frac{4n - 4 - 2a_n}{\pi}\right) = \lim_{n\to \infty} \left(\frac{4 \arctan n}{\pi} - 1\right)^n$$

Now, to compute the latter limit, consider $$\lim_{n \to \infty} \exp \log \left(\frac{4 \arctan n}{\pi} - 1\right)^n = \lim_{n\to \infty} \exp \left(n \log \left(\frac{4 \arctan n}{\pi} - 1\right) \right)$$

Now use L'Hopital on the inside of the $\exp$ function.

  • $\begingroup$ Yes, that's where I got as well. Isn't this equal to $$\lim _{n\to \infty }\left(\frac{\left(4\pi\right)}{\pi }-1\right)^n$$ which would be $3^n$ ? $\endgroup$ – MikhaelM Jul 5 '15 at 14:38
  • $\begingroup$ It's not $3^n$. $\endgroup$ – Zain Patel Jul 5 '15 at 14:42
  • $\begingroup$ @MikhaelM, two things. 1) $$\lim_{n\to\infty} \arctan (n) = \pi /2$$ 2) In general, $$\lim_{n\to\infty} (f(n))^n \not= (\lim_{n\to\infty}f(n))^n$$ $\endgroup$ – FH93 Jul 5 '15 at 14:44
  • $\begingroup$ @FH93 Thanks for the reply. This is where I got stuck. Going to see how to solve the limit now. $\endgroup$ – MikhaelM Jul 5 '15 at 14:47

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