Sum of harmonic series 小學加數

1^ -2 + 2^ -2 + 3^ -2 + 4^ -2 + ...................+ k^ -2 + ..........

=∑ 1/n2

= π2/6

Proof:



(1)


(cos x + i sin x)n = cos(nx) + i sin(nx)



(2)


[cos(nx) + i sin(nx)]/(sin x)n = (cos x + i sin x)n/(sin x)n = [(cos x + i sin x)/sin x]n



(3) we have:



[(cos x + i sin x)/sin x] = cot x + i



(4) Putting this all together gives us:



[ cos(nx) + i sin(nx) ]/(sin x)n = (cot x + i)n



(5) Using the Binomial Theorem


(cot x + i)n = cotn x + (n,1)(cotn-1x)i + ... + (n,n-1)(cot x)in-1 + in



(6) Since i2 = -1, we have:



cotn x + (n,1)(cotn-1x)i + ... + (n,n-1)(cot x)in-1 + in = [(cotnx) - (n,2)(cotn-2 x) ± ... ] + i[(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ]



(7)


[isin(nx)]/(sin x)n = i[(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ]



(8) Dividing both sides by i gives us:



sin(nx)/(sin x)n = [(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ]



(9) Now, since step #8 is an identity, it is true for all values n,x.



Let n = 2m + 1 where m is a positive integer.

Let x = rπ/(2m + 1) where r can = 1, 2, ..., m



From this, we can see that:



nx = (2m + 1)*(rπ)/(2m+1) = rπ



(10)


sin(nx) =0



(11) Applying these values to step #8 gives us:



sin(nx)/(sin x)n = 0/(sin x)n = 0 =



= [(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ] =



= [(2m+1,1)(cot2mx) - (2m+1,3)(cot2m-2x) ± ... + (-1)m ]



(12) we know that the values for x in step #11 are distinct. By the above equation we know that each of these m distinct solutions for x are roots to the following equation:



(2m + 1, 1)tm - (2m + 1,3)tm-1 ± ... + (-1)m = 0



(13) Using Lemma 2 to calculate the sum of roots using the coefficients, we have:



cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) = (2m+1,3)/(2m+1,1) = [(2m+1)!/(2m-2)!(3!)]/[(2m+1)!/(2m)!] = [(2m+1)(2m)(2m-1)/6]/(2m+1)] = (2m)(2m-1)/6.



(14)





cot2x = csc2 x - 1



(15) Applying (cot2x = csc2x - 1) to step #13 gives us:



cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) =



csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1]) - m =



= (2m)(2m-1)/6.



(16) If we add m to both sides we get:



csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1]) = (2m)(2m-1)/6 + m = (2m)(2m-1)/6 + 6m/6 = (4m2 - 2m + 6m)/6 = (4m2 + 4m)/6 = (2m)(2m+2)/6



(17) we know that for the interval (0,π/2) we have the following inequality:



cot2x is less than 1/x2 which is less than csc2 x



(18) But this implies that:



cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) is less than ([2m+1]/π)2 + ([2m+1]/2π)2 + ... + ([2m+1]/mπ)2 which is less than csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1])



(19) Now since we know that



(a) cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) = (2m)(2m-1)/6



(b) csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1]) = (2m)(2m+2)/6



We have:



(2m)(2m-1)/6 is less than ([2m+1]/π)2 + ([2m+1]/2π)2 + ... + ([2m+1]/mπ)2 which is less than (2m)(2m+2)/6



(20) If we multiply on all sides by (π/(2m+1))2, then we have:



π2/(2m+1)2(2m)(2m-1)/6 = (π2)(2m)(2m-1)/(6)(2m+1)2



(π2)/(2m+1)2([2m+1]/π)2 + ([2m+1]/2π)2 + ... + ([2m+1]/mπ)2 = 1/1 + 1/4 + .. + 1/m2



(π2)/(2m+1)2(2m)(2m+2)/6 = π2(2m)(2m-1)/(6)(2m+1)2



(21) as m approaches infinitywe have:



∑ (k=1,∞) (1/k2) = lim (m → ∞) (1/12 + 1/22 + ... + 1/m2) = π2/6



QED

好～