2

If we find the area under the curve of $y=\ln{x}$ from $n$ to $2n$, then will this equal the area of all the rectangles under the curve with width $1$ and height $\ln{k}$ in the same bounds as $n$ approaches infinity?

The rectangles would have areas $\ln{n},\ln{(n+1)},\ln{(n+2)},...,\ln{(2n-2)},\ln{(2n-1)}$

And so the sum of all these areas would be: $$A=\ln{\left(\frac{(2n-1)!}{(n-1)!}\right)}$$

Then we have the integral of $\ln{x}$ from $n$ to $2n$:

$$A=\int_n^{2n}\ln{x} {dx}$$

$$=n\ln{\left(\frac{4n}{e}\right)}$$

Then we have:

$$\lim_{n \to \infty}\ln{\left(\frac{(2n-1)!}{(n-1)!}\right)}=\lim_{n \to \infty}n\ln{\left(\frac{4n}{e}\right)}$$

And then:

$$\lim_{n \to \infty}\left(\frac{(2n-1)!}{(n-1)!}\left(\frac{4n}{e}\right)^{-n}\right)=1$$

Is this right? Am I correct in assuming that the two areas will be equal to each other at the limit?

Sick Nutmeg
  • 977
  • 2
    Both limits are infinite as $n$ approaches infinity. – catherine Oct 30 '22 at 04:49
  • 1
    Do you want to ask if $\lim\limits_{n \to \infty} \frac{f(x)}{g(x)} = 1$ instead of $\lim\limits_{n \to \infty} f(x) = \lim\limits_{n \to \infty}g(x)$ ? – Lab Oct 30 '22 at 05:00
  • The rectangles won't have a width of $1$. If you want to find the exact area, you would need to take the rectangles infinitesimally small, which is basically what a definite integral does. – Accelerator Oct 30 '22 at 05:01
  • @Lab Yes I should probably ask that instead – Sick Nutmeg Oct 30 '22 at 05:01
  • $\lim\limits_{n \to \infty} f(n) = \lim\limits_{n \to \infty}g(n)$ will not always end to $\lim\limits_{n \to \infty} \frac{f(n)}{g(n)} = 1$.

    A simple example is $f(n)=n$ and $g(n)=n^2$.

    You can use Stirling's Formula to show that the last limit is $\frac{\sqrt{2}}{2}$.

     – Henry Oct 30 '22 at 05:54
  • See https://math.stackexchange.com/a/2847768/72031 – Paramanand Singh Nov 01 '22 at 02:49

0 Answers0