Integrals REEEEEEEEEEEEEE

write down x of r in terms of r,
Write down the sum
regognize it's some integral
observe how the sum reads like the line in the end
Obseeve that the integral is oneover a half-cycle or so

>>8970240
watch this video, similar problem

https://www.youtube.com/watch?v=m3NnRTjWuOg&index=21&list=PLj7p5OoL6vGxe7hIWOKfevOet0e1jOezd

>>8970257

Thanks anon, will have a look :D

This question basically tells you the answer....

You just have to see the sum as the Riemann sum of the integral of 1/(1+x^2) from 0 to 1, which is indeed pi/4

To see this just multiply every fraction in the limit by (1/n^2)/(1/n^2) and simplify to see that its f(x) partitioned in [0,1]

You have to funky sum up a bunch of super duper thin rectangles

>>8970375
triggered fuck that stupid bitch that wrote that

>>8970240
Prove that the sum above is equivalent to the limit below. (First do the case n=1 and then induction)

Thus, when you take the limit to infinity of both things, they are equal.

Now work with the sum, turn it into a riemann sum, then convert it into an integral and use the fundamental theorem of calc to compute.

File: 84c37926651114219307e13141c7014f8d3108fe91c9ccf42d666801e8b7047b_1.jpg (108KB, 520x797px) Image search: [Google]

108KB, 520x797px

>>8970240
>comp sci student

>>8970240
The funky sum of 1/(1+x^2) from 0 to 1 is inverse arc reciprocal inverse cosinus over arcsinus done at 1 then subtracted from where it's done at 0.

>>8970257
Yep, you do it this way.
The answer is π/4.