Find the sum of infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...

Question

Find the sum to infinity of the geometric progression:

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots

(NCERT Class 11, Exercise 9.3)

Solution — Step by Step

First term: $a = 1$

Common ratio: $r = \dfrac{1/2}{1} = \dfrac{1}{2}$

Since $|r| = 1/2 < 1$ , the infinite sum converges.

S_\infty = \frac{a}{1 - r}, \quad \text{valid when } |r| < 1

S_\infty = \frac{1}{1 - 1/2} = \frac{1}{1/2} = \mathbf{2}

$S_1 = 1$ , $S_2 = 1.5$ , $S_3 = 1.75$ , $S_4 = 1.875$ , $S_5 = 1.9375$

The partial sums are approaching 2 — consistent with our answer.

Why This Works

Each term adds half of what the previous term added. So the total keeps growing, but by smaller and smaller amounts. The sum never exceeds 2 because you’re always covering half the remaining distance to 2.

Formally, the partial sum $S_n = \frac{a(1 - r^n)}{1 - r}$ . As $n \to \infty$ , $r^n \to 0$ (since $|r| < 1$ ), leaving $S_\infty = \frac{a}{1-r}$ . The condition $|r| < 1$ is essential — if $|r| \geq 1$ , the terms don’t shrink and the sum diverges.

Alternative Method — Algebraic trick

Let $S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$

Multiply both sides by $r = 1/2$ :

\frac{S}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots

Subtract: $S - \frac{S}{2} = 1$ (all other terms cancel)

\frac{S}{2} = 1 \implies S = 2

This subtract-and-cancel trick is the derivation behind the formula itself. Understanding it helps in series problems where the standard formula doesn’t directly apply — like arithmetico-geometric series, where you’ll use a similar shift-and-subtract technique.

Common Mistake

Students sometimes apply $S_\infty = \frac{a}{1-r}$ even when $|r| \geq 1$ . For example, for the series $1 + 2 + 4 + 8 + \cdots$ ( $r = 2$ ), the formula gives $\frac{1}{1-2} = -1$ , which is nonsensical — the sum is clearly growing without bound. Always check $|r| < 1$ before using the infinite sum formula. If $|r| \geq 1$ , the infinite sum simply does not exist.

Question

Solution — Step by Step

Why This Works

Alternative Method — Algebraic trick

Common Mistake

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