Sum of infinite GP = a/(1-r) when |r| < 1. Here a=1, r=1/2. Sum = 2.

Find Sum to Infinity: 1 + 1/2 + 1/4 + ... — Infinite GP Formula

Question

Find the sum to infinity of the series:

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

Solution — Step by Step

The first term is $a = 1$ . To find the common ratio $r$ , divide any term by the one before it:

r = \frac{1/2}{1} = \frac{1}{2}

Check: $\frac{1/4}{1/2} = \frac{1}{2}$ ✓ — consistent throughout, so this is indeed a GP.

The infinite GP formula only works when $|r| < 1$ . Here $|r| = \frac{1}{2} < 1$ , so the series converges — meaning it actually has a finite sum.

This is the most important check. If $|r| \geq 1$ , the terms don’t shrink and the sum blows up to infinity.

The formula for sum to infinity is:

S_\infty = \frac{a}{1 - r}

Substitute $a = 1$ and $r = \frac{1}{2}$ :

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

Why This Works

Each term we add is half the previous one. So we keep adding smaller and smaller amounts — $\frac{1}{2}$ , then $\frac{1}{4}$ , then $\frac{1}{8}$ — and the total creeps closer and closer to 2 but never crosses it.

The formula $S_\infty = \frac{a}{1-r}$ comes from taking the finite sum formula $S_n = \frac{a(1 - r^n)}{1-r}$ and letting $n \to \infty$ . When $|r| < 1$ , the term $r^n \to 0$ , so the formula simplifies to $\frac{a}{1-r}$ .

This is a classic NCERT Class 11 result, and it shows up in JEE Main in more disguised forms — like when $r$ is a fraction involving $x$ , and you’re asked for what range of $x$ the sum exists.

Alternative Method

We can verify this using the partial sum formula and a limit.

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

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

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

S_\infty = \lim_{n \to \infty} 2\left(1 - \frac{1}{2^n}\right) = 2(1 - 0) = 2

Same answer. This approach helps you see why the sum converges — you’re watching $S_n$ inch toward 2 as $n$ grows.

Common Mistake

Many students forget to verify $|r| < 1$ before applying the formula and blindly compute $\frac{a}{1-r}$ for any GP. For example, with the series $1 + 2 + 4 + 8 + \cdots$ , we have $r = 2$ , and the formula gives $\frac{1}{1-2} = -1$ — a negative answer for a series of positive numbers. That’s nonsensical. The sum to infinity simply doesn’t exist here. Always check the convergence condition first.

If you see a series like $1 + x + x^2 + x^3 + \cdots$ in JEE, the sum to infinity is $\frac{1}{1-x}$ for $|x| < 1$ . This is the same formula with $a = 1$ and $r = x$ . You’ll see this pattern frequently in limits and binomial approximation problems too.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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