series問題

General term, 通項
= 2/[(n)(n + 1)]
= 2/[1/n - 1/(n+1)]

該數列的頭n項之和
= 2[1 - 1/2 + 1/2 - 1/3 + ... + 1/n - 1/(n+1)]
= 2[1 - 1/(n+1)]

該數列的無限項之和
= 2(1 - 0) 由於n值為無限, 故1/n+1 = 0]
= 2

無限項之和 :1/(1-1/3)=3/2