The tutor follows the idea of the telescoping sum.

In my previous post, I gave an example of a telescoping sum. It implies that

1-x^(n+1)=(1-x)(1+x+x^2+x^3+x^4+…..+x^n)

Dividing both sides by 1-x we arrive at

(1-x^(n+1))/(1-x)=1+x+x^2+x^3+…..+x^n

which is familiar from the sum of a geometric series of n+1 terms.

If -1<x<1, then x^2<x, x^3<x^2 and so on. Therefore,

lim_n→∞x^(n+1)=0, leading to

(1-x^(n+1))/(1-x)=1/(1-x)=1+x+x^2+x^3+x^4+…. when -1<x<1

Above is the basis for the sum of an infinite series whose common ratio, x, follows -1<x<1.

There is even more to say about the identities shown here; I’ll be following up soon:)

Sources:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Addison-Wesley: Toronto, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.