Prove using PMI (Principle of Mathematical Induction) 1/(1∙2)+1/(2∙3)+1/(3∙4)+⋯+1/n(n+1) =n/(n+1)

Question

Answer 1

There is more simple way: 1/(1·2) + 1/(2·3) + (1/(3·4) + ...+ 1/n(n+1) = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/ 4 + ...+ 1/n -

1/(n + 1) = 1 - 1/(n + 1) = n/(n + 1)

Answer 2

Hi Elizar:

This is how to tackle a question using PMI.

First, we assume that the statement is true for some integer k

=> 1/1. 2 + 1/2 . 3 + . . . + 1/k . (k+1) = k/(k+1) . . . . . . . . . . . eqn 1

Let's write the LHS as S_k

Now, consider the series with one extra term, when k = k+1

We need to show that S_k+1 has exactly the same form as eqn 1 but with k replaced by k+1

Then, S_k+1 = S_k + 1/(k+1)(k+2)

= k/(k+1) + 1/(k+1)(k+2)

= [k(k+2) + 1] / (k+1)(k+2)

= (k²+2k +1) / (k+1)(k+2)

= (k+1)² / (k+1)(k+2)

= (k+1) / (k+2)

= (k+1) / ((k+1)+1)

And this does have the same form as eqn 1 but with k+1 replacing k.

Therefore, if it's true for n=k, it is also true for n=k+1

Now, when n=1, S₁ = 1/1 . 2 = 1/2 = 1/1(1+1) so the statement is true for n=1

Therefore, by the above proof, it must be true for n= 1+1 = 2

and also for n=2+1 =3

and so on, for all integers n.