\sum_1^\infty 1/(n^2*(n+1)^2)
=\sum_1^\infty (1/n-1/(n+1))^2
=\sum_1^\infty 1/n^2+1/(n+1)^2-2*(1/n-1/(n+1))
=\sum_1^\infty(2/n^2)-1-2*\sum_1^\infty(1/n-1/(n+1))
左邊這個級數(shù)是1/n^2,可以用Fourier級數(shù)對f(x)=|x|展開在0點處求得等于π^2/6
右邊這個級數(shù)是1/n/(n+1),后面的項會全消掉,等于1
所以最后=π^2/3-3