1

So i have problems calculating 2 limits :

$$ \lim_{n \to \infty}\frac{\ln(2!)-\ln(n!)-n(\ln(n)))}{n} $$

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt{n^2-k^2}}{n^2}$$

So i have the basic idea that i will somehow need to use partitions. I know at least to get $$\frac{1}{n}$$ out of fraction, however that's where i get stuck. So i know i need to get a set of partitions, then add one element on edge. possibly 0 or 1, and define a map, and then use definite integral with $$[a,b]$$ So basically i need to use Riemann sum to calculate limits. Any hint or even help with one limit would be really appreciated. Thank you in advance. And if there is any similar questions already(possibly duplicates), i'm apologizing for that also. But i did not find any.

janmarqz
  • 10,538
MathIsTheWayOfLife
  • 2,091
  • The second limit is easy. Divide the numerator and denominator by $n$, you get$\sum \frac{\sqrt{1-(k/n)^2}}{n}$ which is $\int_{0}^{1}\sqrt{1-(x)^2}.dx$ – Potemkin Metro Card Dec 03 '15 at 16:28

2 Answers2

2

Hint:

$1)$ $\log n! = \sum_{k=1}^n \log k = \sum_{k=1}^n \log(k/n) - n\log n$

$2)$ $$ \frac{\sqrt{n^2 - k^2}}{n^2} = \frac1{n}\sqrt{\frac{n^2 - k^2}{n^2}} = \frac1{n} \sqrt{1 - (\frac{k}{n})^2}$$

$3)$ $$\lim_{n\to \infty} \frac1{n} \sum_{k=1}^n f(k/n) = \int_0^1 f(x) dx$$

  • So as $$f(x)$$ i can use: $$\sqrt{1-x^2}$$? – MathIsTheWayOfLife Dec 03 '15 at 16:36
  • Yes ${{{{{}}}}}$ –  Dec 03 '15 at 16:38
  • Also D={1/n,....n/n} and C={0,....,n/n} we use D and C as notation first is partitions and second is used as the second set in Riemann sum. If you understand. Also i would appreciate if you could post more detailed for first limit. Thank you. – MathIsTheWayOfLife Dec 03 '15 at 16:41
  • I'm not sure I understood what you said @user246608. Anyway, I think that you can work out the first limit by yourself given my hint. Just write it down on a paper. –  Dec 03 '15 at 16:47
0

All set up now: $$\frac{\sqrt{n^2-k^2}}{n^2} = \frac{\sqrt{n^2(1-(\frac{k}{n})^2)}}{n^2} = \frac{\sqrt{1-(\frac{k}{n})^2}}{n} $$

Victor
  • 3,213