Answer :
Step-by-step explanation:
let two successive integers n, n+1
difference of the reciprocals:
[tex]\frac{1}{n}-\frac{1}{n+1} \\=\frac{(n+1)-n}{(n)(n+1)}\\=\frac{1}{(n)(n+1)}\\=\frac{1}{(n)}\frac{1}{(n+1)}\\[/tex]
Step-by-step explanation:
let two successive integers n, n+1
difference of the reciprocals:
[tex]\frac{1}{n}-\frac{1}{n+1} \\=\frac{(n+1)-n}{(n)(n+1)}\\=\frac{1}{(n)(n+1)}\\=\frac{1}{(n)}\frac{1}{(n+1)}\\[/tex]