Why the $\lim_{n\to\infty} (\frac{n}{n^2+1}+\frac{n}{n^2+4}+.....+\frac{n}{n^2+n^2})= \frac{\pi}{4}$?
I read somewhere that it is related to $f(x)=\frac{1}{1+x^2}$ but dont know why...
Asked
Active
Viewed 151 times
1
-
1Think of Riemann sums. – Clayton Oct 10 '18 at 00:11
2 Answers
3
Use the def of Riemann integration. Write
$$ \lim_{n \to \infty} \sum_{i=1}^n \frac{ n}{n^2+ i^2} = \lim_{n \to \infty} \sum_{1}^n \frac{1/n}{1+(i/n)^2} = \int\limits_0^1 \frac{1}{1+x^2} dx = \frac{\pi}{4} $$
-
why $\frac{1}{n}$ in the sum became 1 in the integral? how you transfer that? – sam0101 Oct 10 '18 at 00:20
-
1@sam0101 It should really be written as $\lim_{n\to\infty}{\frac1n}\sum_{i=1}^n \frac1{1+(i/n)^2}$; the first factor of $1/n$ amounts to the 'dx' term, the partition size for the Riemann integral. – Steven Stadnicki Oct 10 '18 at 00:22
-
2
$$\sum_{i=1}^n \dfrac{n}{n^2+i^2}=\sum_{i=1}^n \dfrac{n^2}{n^2+i^2}\dfrac{1}{n}=\sum_{i=1}^n \dfrac{1}{1+(0+i/n)^2}\dfrac{1}{n}$$
Now reimagine: $\Delta x = \dfrac{1-0}{n}, x_i = 0+i\Delta x = \dfrac{i}{n}$
Then your sum is equal to
$$\sum_{i=1}^{n} f(x_i) \Delta x$$
Now take the limit as $n\to\infty$
David P
- 12,320