Consider the finite geometric series [math]S= a+ at+ at^2+ at^3+ \cdot\cdot\cdot+ at^n[/math]. Subtract a from both sides: [math]S- a=at+ at^2+ at^3+ \cdot\cdot\cdot+ at^n[/math]. Factor "t" out of the right side: [math]S- a= t(a+ at+ at^2+ \cdot\cdot\cdot+ at^{n- 1})[/math]. The sum on the right is almost the same as the original sum- it is only missing the "[math]at^n[/math]" term. Add and subtract [math]at^n[/math] inside the parentheses:
[math]S- a= t(a+ at+ at^2+ \cdot\cdot\cdot+ at^{n- 1}+ at^n- at^n)= t(a+ at+ at^2+ \cdot\cdot\cdot+ at^{n- 1}+ at^n)- at^{n+1}= tS- at^{n+1}[/math]
Add [math]at^{n+1}[/math] to both sides:
[math]S- a+ at^{n+ 1}= tS[/math]
[math]S- tS= (1- t)S= a- at^{n+1}= a(1- t^{n+1})[/math]
Finally, divide both sides by 1- t:
[math]S= \frac{a(1- t^{n+1})}{1- t}[/math]
The sum of the infinite series, [math]\sum_{n=0}^\infty at^n[/math] is the limit of that as n goes to infinity. If [math]t\ge 1[/math], that limit, so that sum, does not exist. If [math]t<1[/math], [math]t^{n+ 1}[/math] goes to 0 so
[math]\sum_{n=0}^\infty at^n= \frac{a}{1- t}[/math].
