I spent the night trying to test a theory of mine (irrelevant), which brought me to this limit:

EDIT: Here's a picture, much easier:



I expect the answer to be π^-0.5 (≈ 0.564), but my approximations up to 10000 iterations are closer to 0.574, off by an entire hundredth. I need a way to compute the exact limit as N approaches infinity, but I can't seem to simplify the summation any further than I already have (which is an improvement on the original).

I know that plenty of you must have taken some calculus - maybe you know the next step here. Any ideas?

Note: Since this is not a problem from any textbook, it may very well be the case that this cannot be simplified. But, I simply refuse to believe that right now.

Do you honestly expect an answer from here?

I could ask a room mate, but you're better off asking your professor if this is a theory (and presumably not homework related).

I did look at this earlier since I have taken all the way through Calc III (Vector Calc)... but I always hated limits and found that to be the most difficult part of Calculus for me. So I had no idea what to do for this problem, sorry.

While I've had Calculus 1-3, that's also been like 9 years ago. I had Calc 1 in high school, and 2 & 3 in college. Since my job doesn't encourage daily calculus problems, I really don't remember much about it. Sorry.

I asked a mathematician friend of mine. I had come to the correct limit, but it wasn't what I wanted it to be. That 0.57... is correct.



It was worth a shot :p

This was an attempt to construct a circle with the same area as a unit square. I know that the radius of such a circle must be 1/sqrt(pi), so I was a little dismayed when my limit didn't match my expected answer.

I need to find out why the answer, for any size square, is always close to the expected answer but slightly off. Surely the limit with this method approaches the correct radius length as the number of sides in a regular polygon approach infinity, so there must be done very simple adjustment that I can make to apply this to a square...