Q.A substance initially weighing 64 grams is decaying at a rate such that after 8 hours there are only 32 grams remaining. In another 4 hours there are only 16 grams left, and after 2 more hours only 8 grams remain. How much time passes before none of the substance remains?

Infinite Geometric Progression

The intuition: what happens when you keep halving?

Imagine you start walking toward a wall. In the first step, you cover half the remaining distance. In the next step, you cover half of what's left. Then half of that. And so on.

The distances you cover in each step are:

$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$

This is a geometric progression — each term is the previous term multiplied by a fixed number (here, $\frac{1}{2}$). That fixed multiplier is called the common ratio, $r$.

Now ask: if you keep doing this forever, what is the total distance you cover? Intuition says you never quite reach the wall — but you get arbitrarily close. The total distance covered after $n$ steps is:

$S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n}$

This sum gets closer and closer to $1$ as $n$ grows. It never exceeds $1$, but it approaches $1$ as closely as you like. We say the infinite sum equals $1$.

That is the core idea of an infinite geometric progression: a GP that goes on forever, and whose sum (under the right condition) settles to a finite number.

The precise statement

An infinite geometric progression is a sequence of the form:

$a, ar, ar^2, ar^3, ar^4, \dots$

where $a$ is the first term and $r$ is the common ratio. The sequence never ends.

The sum of the first $n$ terms (the finite sum) is:

$S_n = a \frac{1 - r^n}{1 - r}, \quad r \neq 1$

Now, what happens as $n \to \infty$? That depends entirely on $r$.

Why $|r| < 1$ is the key

Look at $S_n = a \frac{1 - r^n}{1 - r}$. The only part that depends on $n$ is $r^n$.

• If $|r| < 1$, then $r^n \to 0$ as $n \to \infty$. So $S_n \to \frac{a}{1 - r}$.
• If $|r| > 1$, then $r^n$ grows without bound — the sum diverges.
• If $r = 1$, the formula breaks (division by zero), and the sum is $a + a + a + \dots$, which diverges.
• If $r = -1$, the terms alternate $a, -a, a, -a, \dots$; the partial sums bounce between $a$ and $0$ and never settle.

A quick example

Problem: Find the sum $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$

Here $a = 1$, $r = \frac{1}{3}$. Since $|r| = \frac{1}{3} < 1$, the sum exists:

$S_\infty = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

So the infinite sum is $\frac{3}{2}$.

The big picture

An infinite GP is not magic — it is just a GP that keeps going. The only new idea is that when the common ratio is small enough (between $-1$ and $1$), the later terms become so tiny that the total sum converges to a finite number. That number is given by $\frac{a}{1-r}$.

Every time you see an infinite geometric series, your first two questions should be: What is $r$? Is $|r| < 1$?