Recent content by bank

I found my required answer simple enough. I could manipulate my series to (2/pi).SUM(from n=1 to infinity) of (-1)^n+1 . pi^2n / (2n)! This is thus after a change of sign: -(2/pi) [pi^2/2! - pi^4/4! + pi^6/6! -...] =-(2/pi) . cos(pi) - 1 = 4/pi as required I've had a look next week, and...

I seem to be getting somewhere now. I'd still like to know how to do it the way you've suggested using definite integrals, but by using the strategy I was thinking of, that is expanding the sines as series and substituting the values for the integer power summations, I arrive at an answer for...

Have you got any examples of using definite integrals to solve this sort of thing?

Cheers, I'll have to have a look later, do you mean the whole thing, or just the part after the summation? I'll have to check my notes as I haven't done any analysis in a year. Until you mentioned that I was going to look into the sums for: 1+2+3+4+5+...+(n-1) = ? 1^3+2^3+3^3+...+(n-1)^3...

Homework Statement *Find the distance between 1 and the various n-th roots of unity - denoted d(k) *Find a formula for the sum of distances between 1 and each of the n-th roots of unity - denoted S(n) *Find the limit as n->infinity of (1/n).S(n) Homework Equations *The n-th roots...