Find sum to infinity of 1/2 + 1/4 + 1/8 + ...

Question

Find the sum to infinity of the series: $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots$

Solution — Step by Step

Write out the terms: $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots$

Check the ratio between consecutive terms:

\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}, \quad \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2}

The common ratio $r = \frac{1}{2}$ is constant. This is a Geometric Progression with:

First term $a = \frac{1}{2}$
Common ratio $r = \frac{1}{2}$

An infinite GP has a finite sum (converges) if and only if $|r| < 1$ .

Here $|r| = \frac{1}{2} < 1$ . ✓

As we add more and more terms, each new term $\left(\frac{1}{2}\right)^n \to 0$ as $n \to \infty$ . The sum approaches a finite limit.

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

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

The sum to infinity = 1.

This is one of the most beautiful results in mathematics: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = 1$ .

Think of it physically: start with a 1-metre ribbon. Cut off half (½ m). From the remaining half, cut off half again (¼ m). Keep going. You’re always adding more ribbon — but you never exceed 1 m total, and you approach exactly 1 m.

The partial sums are: $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots$ — each time you’re halfway between the current sum and 1. The limit is 1.

Why This Works

The derivation of the formula is elegant. Let $S = a + ar + ar^2 + ar^3 + \ldots$

Multiply both sides by $r$ : $rS = ar + ar^2 + ar^3 + \ldots$

Subtract: $S - rS = a$ (all other terms cancel)

S(1 - r) = a \implies S = \frac{a}{1-r}

This derivation assumes $|r| < 1$ , which ensures the terms approach 0 and the series doesn’t diverge.

If $|r| \geq 1$ , terms don’t approach 0 — the sum is infinite (or doesn’t converge). For example: $1 + 2 + 4 + 8 + \ldots$ has $r = 2$ and clearly diverges to infinity.

Alternative Method — Telescoping Partial Sums

The sum of the first $n$ terms of a GP:

S_n = \frac{a(1 - r^n)}{1 - r} = \frac{\frac{1}{2}(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n}

As $n \to \infty$ : $\frac{1}{2^n} \to 0$ , so $S_n \to 1$ .

This confirms the limit from first principles without needing the sum-to-infinity formula directly.

CBSE Class 11 and JEE Main both test this formula. Watch out for questions that ask “find the sum to infinity of $3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots$ ” — here the series alternates in sign, meaning $r = -\frac{1}{3}$ , and $|r| = \frac{1}{3} < 1$ so the formula still applies: $S = \frac{3}{1 - (-\frac{1}{3})} = \frac{3}{\frac{4}{3}} = \frac{9}{4}$ .

Common Mistake

Students sometimes apply the sum-to-infinity formula when $|r| \geq 1$ . For example: $1 + 2 + 4 + 8 + \ldots$ — if you blindly apply $\frac{a}{1-r} = \frac{1}{1-2} = -1$ , you get $-1$ , which is nonsensical (a sum of positive numbers can’t be negative). Always check $|r| < 1$ before applying the formula. A sum of positive increasing terms cannot have a finite sum.

Question

Solution — Step by Step

Why This Works

Alternative Method — Telescoping Partial Sums

Common Mistake

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