Question

A.Find a formula for

Answer 1

a. Notice that

1/(1*2) = 1/2 = 1 - 1/2

1/(2*3) = 1/6 = 1/2 - 1/3

1/(3*4) = 1/12 = 1/3 - 1/4

and so on, which suggests the n-th term of the sum can be written as

$\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$

Then the sum itself is telescoping:

$\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{n(n+1)}$

$=\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\left(\dfrac13-\dfrac14\right)+\cdots+\left(\dfrac1n-\dfrac1{n+1}\right)$

$=1-\dfrac1{n+1}$

b. The proof is trivial:

$\dfrac1n-\dfrac1{n+1}=\dfrac{n+1}{n(n+1)}-\dfrac n{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\dfrac n{n+1}$

so the formula found in (a) is correct.