Convergence of Geometric Series With Ratios Below One

Why a Geometric Series Can Have a Finite Sum

A geometric series continues indefinitely, but its terms can become small quickly enough that its partial sums approach a finite number. For instance, the positive terms \(4, 2, 1, \frac12,\ldots\) keep adding to the total, while each new term is half the previous one. The resulting sum is finite.

The key condition is that the common ratio \(r\) satisfies \(|r|<1\). This includes negative ratios, which produce alternating signs. In this tutorial, we use the finite partial-sum formula to prove convergence and calculate the sum.

We use the indexing convention that the first term is \(a\), corresponding to \(n=0\). The common ratio is the signed number \(r\): each term is obtained by multiplying the preceding term by \(r\). Thus a negative ratio means the signs alternate.

The Finite Partial-Sum Formula

For \(r\ne 1\), the finite partial sum is

Here is a direct derivation. Multiply the sum by \(r\), then subtract the resulting expression from the original sum. All the intermediate terms cancel:

Dividing by \(1-r\) gives the formula. When \(|r|<1\), we have \(r\ne 1\), so this division is valid.

Why the Powers Approach Zero

To take the limit of the partial-sum formula, we need to know that \(r^{N+1}\) approaches zero when \(|r|<1\). For a negative ratio, the powers alternate in sign, so consider their absolute values: \(|r^{N+1}|=|r|^{N+1}\).

Let \(q=|r|\). If \(q=0\), then every positive power of \(q\) is zero. If \(0<q<1\), set \(h=(1-q)/q\), so \(h>0\) and \(1/q=1+h\). For every positive integer \(m\),

This inequality follows by induction: it holds for \(m=1\); and if it holds for \(m\), then

Taking reciprocals of positive quantities gives

As \(m\) increases, \(1/(1+mh)\) approaches zero: for any \(\varepsilon>0\), choosing \(m\) large enough that \(1+mh>1/\varepsilon\) makes \(1/(1+mh)<\varepsilon\). The displayed bound then shows that \(q^m\) also approaches zero. Therefore \(r^m\) approaches zero, whether \(r\) is positive or negative.

The Convergence Theorem

The proof follows by taking the limit of the finite partial sums. Since \(r^{N+1}\to 0\), the finite-sum formula gives

The denominator \(1-r\) is nonzero because \(|r|<1\) rules out \(r=1\). The partial sums therefore approach a finite limit, which proves both convergence and the stated value of the sum. This argument works for negative \(r\) as well: the finite partial sums may alternate around their limit, but the term involving \(r^{N+1}\) still approaches zero.

Worked Examples

Absolute Convergence and the Role of the Ratio

When \(|r|<1\), the absolute values of the terms form another geometric series:

Applying the same theorem with first term \(|a|\) and ratio \(|r|\) shows that this series has sum \(|a|/(1-|r|)\), which is finite. Thus the original series is absolutely convergent: the series of absolute values converges. In particular, alternating signs do not prevent convergence under the stated condition.

For the value of the original signed series, however, use \(a/(1-r)\), not \(a/(1-|r|)\). The absolute ratio is used to check the condition for convergence; the signed ratio is used to calculate the sum.

Common Mistake: Replacing the Signed Ratio

For \(6-3+\frac32-\cdots\), it is correct to use \(|r|=\frac12\) to establish convergence. But replacing \(r\) by \(|r|\) in the sum formula gives the wrong answer:

The series actually has signed ratio \(r=-\frac12\). The correct calculation is

The check confirms the correction: \(6+(-\frac12)(4)=4\), whereas \(6+(-\frac12)(12)=0\ne12\). Keep the two roles distinct: \(|r|<1\) is the convergence condition, and \(r\) itself determines the sum.

Using the Result Reliably

To evaluate a geometric series of this form, first identify its first term and signed common ratio. Confirm that the same ratio takes each term to the next, then check \(|r|<1\). Only after that check should you use \(a/(1-r)\). The condition matters because the proof depends on the powers \(r^{N+1}\) approaching zero; without that decay, the limiting step used here is not justified.

Finally, check the arithmetic and the sign. A useful consistency check is that a convergent geometric sum satisfies \(S=a+rS\): separating the first term leaves a series whose terms are \(r\) times the original terms. Solving this equation gives \((1-r)S=a\), hence \(S=a/(1-r)\). This check is not a substitute for verifying the convergence condition; it is a way to catch errors after that condition has been established.