Proving that the sum of a geometric series is of the order of its last term

Question

How can we prove that the sum of a geometric series is of the order of the last term for r>1, that is to show that its theta notation consists of the last term.

Answer 1

For $a=1$ the statement is $$na=O(n)$$ which is true.

Otherwese, $$a+ar+ar^2+...+ar^n = \frac {a(r^{n+1}-1)}{r-1} =O(r^n)$$