【摘要】 对于sum from n=1 to ∞ 1/n^2m（m∈Z⁺）,当n-1时,有sum from n=1 to ∞ 1/n²=π²/6,并且对它有着许多种不同的证法.通过博里叶（Fourier）级数以及逐项积分,得到关于sum from n=1 to ∞ 1/n^2m（m∈Z⁺）的和的系数的一个递推关系式,并给出当m=1,2,3,4,5时的结果。更多还原

【Abstract】 For when m =1,we know ,and there are many proofs.In this paper,with Fourier series and term by term integration, we get a recurrence formula of coefficient about thengives the results of when m = 1,2 ,3,4,5.更多还原