Boundary Geometric Series With Ratios One and Minus One

Introduction

A geometric series is built by repeatedly multiplying each term by the same common ratio. When the ratio has absolute value less than one, its terms shrink, and the partial sums approach a finite limit. At the boundary ratios \(1\) and \(-1\), however, the terms do not shrink unless the first term is zero.

These two cases are useful because their partial sums can be computed exactly. The ratio \(1\) produces a constant term repeated indefinitely. The ratio \(-1\) produces terms that alternate between a number and its negative. In both cases, examining the partial sums makes the convergence question precise.

Series, Terms, and Partial Sums

Let \(a\) be the first term and \(r\) the common ratio. The geometric series is

a+ar+ar^2+ar^3+\cdots = \sum_{k=0}^{\infty} ar^k.

The \(N\)th partial sum is the sum of the first \(N\) terms. Since \(N\) counts terms, it is a positive integer.

S_N=\sum_{k=0}^{N-1} ar^k =a+ar+ar^2+\cdots+ar^{N-1}.

Definition: The infinite series \(\sum_{k=0}^{\infty} ar^k\) converges if the sequence of partial sums \((S_N)\) approaches a finite real number as \(N\) tends to infinity. If the partial sums do not approach a finite limit, the series diverges.

A necessary condition for a series to converge is that its terms approach zero. Here the \(k\)th term is \(ar^k\). This condition can immediately rule out convergence when the terms fail to approach zero. It is still important to inspect partial sums: their behavior is the definition of series convergence.

When the Ratio Is One

If \(r=1\), multiplying by the ratio does not change any term. The series is

a+a+a+a+\cdots.

Every one of its first \(N\) terms is \(a\), so its partial sum is

S_N=\underbrace{a+a+\cdots+a}_{N\text{ terms}}=Na.

If \(a\ne 0\), the magnitude of the partial sum is \(|S_N|=N|a|\), which grows without bound as \(N\) increases. Thus the series diverges. Its terms are all \(a\), so their limit is also the nonzero number \(a\), in agreement with the necessary condition for convergence. If \(a=0\), every term and every partial sum is zero, and the series converges to zero.

Result: For common ratio \(r=1\), the geometric series converges exactly when \(a=0\). In that case its sum is zero; for \(a\ne0\), its partial sums \(S_N=Na\) do not have a finite limit.

When the Ratio Is Minus One

If \(r=-1\), the terms alternate between \(a\) and \(-a\):

a-a+a-a+\cdots.

To determine the partial sum, write out its terms and also write the sum after shifting each term by one power of \(-1\). This shifted sum equals \(-S_N\):

\begin{aligned} S_N&=a+a(-1)+a(-1)^2+\cdots+a(-1)^{N-1},\\ -S_N&=a(-1)+a(-1)^2+\cdots+a(-1)^N. \end{aligned}

Subtract the second equation from the first. Every term between the endpoints cancels, leaving the first term minus the final term of the shifted sum:

$$ S_N-(-S_N)=a-a(-1)^N. $$

Therefore \(2S_N=a(1-(-1)^N)\), and dividing by \(2\) gives the partial-sum formula.

S_N=\frac{a\bigl(1-(-1)^N\bigr)}{2}.

This formula shows exactly what happens for each parity of \(N\). If \(N\) is even, then \((-1)^N=1\), so \(S_N=0\). If \(N\) is odd, then \((-1)^N=-1\), so \(S_N=a\). Thus the partial sums alternate between \(a\) and \(0\).

When \(a\ne0\), the even-indexed partial sums are all zero, while the odd-indexed partial sums are all \(a\). These two subsequences have different limits, so the full sequence of partial sums has no limit. The terms themselves also alternate between \(a\) and \(-a\), and do not approach zero. When \(a=0\), every term and partial sum is zero, so the series converges.

Result: For common ratio \(r=-1\), the series converges exactly when \(a=0\). Its partial sums satisfy \(S_N=0\) for even \(N\) and \(S_N=a\) for odd \(N\); if \(a\ne0\), these partial sums do not converge.

Worked Examples

Worked Example: A Series With Ratio One

Determine whether \(3+3+3+\cdots\) converges, and find the first four partial sums.

The first term is \(a=3\), and the common ratio is \(r=1\). The formula for this boundary case is \(S_N=Na\), so

S_1=1(3)=3,\qquad S_2=2(3)=6,\qquad S_3=3(3)=9,\qquad S_4=4(3)=12.

Directly adding the terms gives the same values: \(3\), \(3+3=6\), \(3+3+3=9\), and \(3+3+3+3=12\). In general \(S_N=3N\), which grows without bound as \(N\) increases. The terms are all \(3\), so they do not approach zero either. The series diverges.

Worked Example: An Alternating Series With Ratio Minus One

Find the first five partial sums of \(6-6+6-6+\cdots\), and determine whether the series converges.

Here \(a=6\) and \(r=-1\). Use \(S_N=\frac{a(1-(-1)^N)}{2}\). For \(N=1,2,3,4,5\), the values are

\begin{aligned} S_1&=\frac{6(1-(-1))}{2}=6,\\ S_2&=\frac{6(1-1)}{2}=0,\\ S_3&=\frac{6(1-(-1))}{2}=6,\\ S_4&=\frac{6(1-1)}{2}=0,\\ S_5&=\frac{6(1-(-1))}{2}=6. \end{aligned}

Checking by direct addition gives \(6\), \(6-6=0\), \(6-6+6=6\), \(6-6+6-6=0\), and \(6-6+6-6+6=6\), as in the formula. The even partial sums remain zero, while the odd partial sums remain six. Since these two subsequences have different limits, the partial sums do not converge. The series diverges.

Worked Example: The Zero First Term

Classify the geometric series with first term \(a=0\) and common ratio \(r=-1\).

Its \(k\)th term is \(ar^k=0(-1)^k=0\), for every \(k\). Consequently, for every positive integer \(N\),

S_N=\sum_{k=0}^{N-1}0(-1)^k=0.

The partial sums therefore have limit zero, so the series converges to zero. The general formula confirms this: \(S_N=\frac{0(1-(-1)^N)}{2}=0\), for even and odd \(N\). Substitution into the original series also gives \(0+0+0+\cdots=0\). This is the exceptional convergent case for ratio \(-1\); the same zero-first-term exception applies when the ratio is \(1\).

A Common Pitfall: Misusing the Finite Geometric Formula

For a finite geometric sum with ratio \(r\ne1\), the usual formula is \(S_N=\frac{a(1-r^N)}{1-r}\). The condition \(r\ne1\) matters because the denominator would be zero when \(r=1\). For example, substituting \(a=3\), \(r=1\), and \(N=2\) produces

\frac{3(1-1^2)}{1-1}=\frac{0}{0},

which is undefined, not a value for the sum. The actual two-term sum is \(3+3=6\), and \(S_N=3N\) gives \(S_2=6\). The ratio-one case must be handled directly rather than by dividing by \(1-r\).

Another error is to infer convergence merely because the partial sums are bounded. For \(6-6+6-6+\cdots\), every partial sum is either \(0\) or \(6\), so the partial sums are bounded. But boundedness does not establish that they approach one value: the even partial sums are \(0\), and the odd partial sums are \(6\). Convergence requires a single finite limit.

Why the Boundary Cases Matter

These examples separate three ideas that are easy to confuse: terms may fail to approach zero, partial sums may fail to approach a limit, and partial sums may remain bounded without converging. For ratio \(1\) with \(a\ne0\), the terms stay fixed and the partial sums grow in magnitude. For ratio \(-1\) with \(a\ne0\), the terms oscillate without approaching zero, and the partial sums oscillate between two distinct values. In both cases, the series diverges.

The zero-first-term case is worth stating explicitly. When \(a=0\), multiplying by either boundary ratio still gives zero at every step. The resulting series is not an exception to the partial-sum definition: its partial sums really do approach a finite limit, namely zero.

Takeaway: For either boundary ratio \(r=1\) or \(r=-1\), the geometric series converges if \(a=0\) and diverges if \(a\ne0\). For \(r=1\), use \(S_N=Na\). For \(r=-1\), use \(S_N=\frac{a(1-(-1)^N)}{2}\) and compare the even and odd partial sums.

Check Your Understanding

Use the partial sums, not just a few terms, to justify each conclusion.

For \(4+4+4+\cdots\), write a formula for \(S_N\) and decide whether the series converges.
For \(5-5+5-5+\cdots\), find \(S_6\) and \(S_7\). What do these values show about convergence?
Why is the finite geometric-sum formula with denominator \(1-r\) not valid at \(r=1\)?
For a geometric series with \(a=0\) and \(r=1\), what are all its terms and its sum?
Can bounded partial sums alone prove that a series converges? Use the ratio \(-1\) case to explain.