Negative Ratios and Oscillating Geometric Sums

Why Negative Ratios Need Careful Tracking

A geometric series with a negative common ratio alternates in sign: positive terms are followed by negative terms, or negative terms by positive terms. This sign change affects the partial sums, but it does not prevent them from converging when the ratio’s magnitude is less than one.

To understand what the sums do, keep two questions separate. First, what is the exact value of a finite partial sum? Second, what happens to those partial sums as the number of terms grows? The finite formula answers the first question; its limiting behavior answers the second.

Terms and Partial Sums

Let \(a\) be the first term and \(r\) the common ratio. The \(k\)-th term, with the index beginning at \(k=0\), is \(ar^k\). Thus each term after the first is obtained by multiplying the preceding term by \(r\).

For a positive integer \(n\), define \(S_n\) to be the sum of the first \(n\) terms. With this indexing, the first \(n\) terms have exponents from \(0\) through \(n-1\).

In this tutorial, the ratio is negative, \(r<0\), and its magnitude is less than one, \(|r|<1\). The first condition causes sign changes for nonzero terms. The second causes the magnitudes of nonzero terms to decrease toward zero. If \(a=0\), however, every term is zero, whatever the value of \(r\); there are no nonzero terms whose signs alternate.

The Finite Sum Formula

When \(r\ne 1\), the first \(n\) terms have the finite geometric sum

For the negative ratios considered here, \(r<0\), so \(1-r\ne 0\) and the formula applies. The negative sign in \(r\) must remain in the powers \(r^n\). Replacing \(r\) by \(|r|\) would change the terms and the sum.

Why the Formula Works

The formula follows by multiplying the finite sum by \(r\) and subtracting. This cancellation argument also shows why the exponent in the last term is \(n-1\).

This derivation is valid for every positive integer \(n\) and every \(r\ne1\). It does not depend on whether \(r\) is positive or negative, or on whether the series has a limit. Here, it gives each finite partial sum exactly.

Worked Example: Five Terms with Ratio \(-\frac12\)

Why the Partial Sums Converge

For \(r<0\) with \(|r|<1\), the powers \(r^n\) alternate in sign, but their magnitudes approach zero. One way to justify the shrinking magnitudes is to set \(q=|r|\), where \(0<q<1\). Then \(1/q=1+c\) for some \(c>0\). The binomial inequality \((1+c)^n\ge 1+nc\) gives

The right-hand side approaches zero as \(n\) grows, so \(q^n\to0\), and therefore \(r^n\to0\). The case \(r=0\) also has \(r^n=0\) for every positive \(n\).

Now apply this fact to the finite sum formula. Since \(1-r\) is a fixed nonzero number, the term involving \(r^n\) vanishes in the limit:

Because \(r<0\), the denominator \(1-r\) is greater than one. The limit is not found by changing the ratio to its magnitude; it comes from using the signed ratio in the formula.

Worked Example: Four Terms with Ratio \(-\frac23\)

Alternation Around the Limit

The partial sums themselves also have a useful relation to their limit \(L=\frac{a}{1-r}\). Subtracting \(L\) from the finite sum formula gives

For \(a\ne0\), the sign of this difference alternates with \(n\), because \(r^n\) does. Thus successive partial sums lie on opposite sides of the limit. Their distance from it is

which tends to zero. If \(a>0\), odd-indexed partial sums lie above the limit and even-indexed ones below it. If \(a<0\), the sides are reversed. If \(a=0\), all partial sums equal the limit, zero, and there is no oscillation around it.

The change from one partial sum to the next is exactly the next term:

So for nonzero \(a\), the partial sums move alternately upward and downward, with steps whose magnitudes shrink. This is why alternating terms do not mean the partial sums fail to settle: the steps become small enough that the sums approach a fixed value.

Worked Example: A Negative First Term

Common Mistake: Replacing the Ratio by Its Magnitude

A frequent error is to use \(|r|\) in the sum formula because it describes the sizes of the terms. But \(|r|\) does not preserve their signs. For example, with \(a=4\) and \(r=-\frac12\), the correct infinite sum is

Replacing \(r\) by \(\frac12\) instead would give \(\frac{4}{1-\frac12}=8\), which is the sum for a different series, \(4+2+1+\cdots\). The actual series is \(4-2+1-\frac12+\cdots\), so its signs must remain in the computation.

There is a related qualification about term magnitudes. If \(a\ne0\) and \(|r|<1\), then each successive term has smaller magnitude, since \(|ar^{k+1}|=|r|\,|ar^k|<|ar^k|\). When \(a=0\), every term has magnitude zero; the magnitudes are equal, not strictly decreasing. Keeping this special case explicit avoids claiming that zero terms shrink.

Why This Topic Matters

Negative ratios show that convergence is not the same as monotone movement. For a nonzero first term, these partial sums alternate around their limit rather than approaching it from just one side. The exact formula makes that behavior visible: \(S_n-L=-\frac{ar^n}{1-r}\) records both the alternating side and the shrinking distance.

When computing, first identify the signed ratio, then use the finite formula with the correct number of terms. Only after that should you take a limit, and only when \(|r|<1\). In particular, the limiting sum is a value approached by the partial sums; it is not generally equal to any one finite partial sum.