Geometric Series at Unit-Magnitude Ratios

Why Ratios of One and Negative One Need Separate Attention

A geometric series is built from terms that are obtained by repeatedly multiplying by the same common ratio. When the ratio has magnitude less than one, the terms may become small; when the ratio is one or negative one, their magnitudes do not decrease. The partial sums make the consequences especially clear.

The key question is not whether the terms look regular, but whether the sequence of partial sums approaches a finite real number. For these two ratios, we can answer that question directly by listing the terms and computing the partial sums.

Convention: Let $a$ be the first term and $r$ the common ratio. We index terms from $k=0$, so the $k$-th term is $a r^k$. The partial sum $S_N$ includes the terms from $k=0$ through $k=N$, giving $N+1$ terms.

Thus the series under consideration is $a+ar+ar^2+\cdots$, and its partial sums are $S_N=\sum_{k=0}^{N}ar^k$. By definition, this infinite series converges to a finite sum exactly when the sequence $(S_N)$ has a finite limit as $N$ increases.

Common Ratio One: Constant Terms

When $r=1$, every term equals $a$, because $a(1)^k=a$ for every nonnegative integer $k$. Adding $N+1$ copies of $a$ gives the partial sum directly.

S_N=\sum_{k=0}^{N}a(1)^k =\underbrace{a+a+\cdots+a}_{N+1\text{ terms}} =(N+1)a.

If $a=0$, every term and every partial sum is zero, so the series converges to zero. If $a\ne0$, the magnitude of the partial sum is $(N+1)|a|$, which grows without bound. In particular, the partial sums cannot approach a finite real number.

Worked Example: A Constant Positive Term

Consider the geometric series with first term $a=3$ and ratio $r=1$. Its first terms are $3,3,3,3,\ldots$. Since $S_N$ includes $N+1$ terms,

$$ S_N=(N+1)3=3N+3. $$

For example, $S_0=3$, $S_1=6$, and $S_3=12$. The formula gives those values: $3(0)+3=3$, $3(1)+3=6$, and $3(3)+3=12$. As $N$ increases, $3N+3$ increases without bound, so the series does not converge to a finite sum.

The partial-sum formula also checks against the terms: for $N\ge1$, $S_N-S_{N-1}=(3N+3)-(3(N-1)+3)=3$, which is the term added at step $N$.

Worked Example: A Constant Negative Term

Now let $a=-2$ and $r=1$. Each term is $-2$, so

$$ S_N=(N+1)(-2)=-2N-2. $$

The first partial sums are $S_0=-2$, $S_1=-4$, and $S_2=-6$. Substitution into the formula gives $-2(0)-2=-2$, $-2(1)-2=-4$, and $-2(2)-2=-6$. The partial sums decrease without bound, and therefore do not have a finite limit. Checking consecutive partial sums confirms that each new term is $-2$: $S_N-S_{N-1}=(-2N-2)-(-2(N-1)-2)=-2$.

Common Ratio Negative One: Alternating Terms

When $r=-1$, multiplication by the ratio changes the sign of each term. The terms alternate between $a$ and $-a$:

a,\ -a,\ a,\ -a,\ \ldots

To determine convergence, pair the terms in the partial sums. Each adjacent pair $a+(-a)$ contributes zero. Whether one term remains depends on whether $N$ is even or odd, because $S_N$ contains $N+1$ terms.

Partial-sum formula: For the geometric series with first term $a$ and ratio $-1$, $$ S_N= \begin{cases} a, & \text{if }N\text{ is even},\\ 0, & \text{if }N\text{ is odd}. \end{cases} $$

For even $N$, the number $N+1$ of terms is odd. All but the last term cancel in pairs, leaving $a$. For odd $N$, there are an even number of terms, and they all cancel in pairs, leaving zero. This gives the formula from the actual terms, without relying on a limiting shortcut.

Worked Example: Alternating Terms Starting at Four

Take $a=4$ and $r=-1$. The terms begin $4,-4,4,-4,\ldots$. The first partial sums are

S_0=4,\qquad S_1=4-4=0,\qquad S_2=4-4+4=4,\qquad S_3=4-4+4-4=0.

The parity formula gives $S_0=4$ because zero is even, $S_1=0$ because one is odd, $S_2=4$ because two is even, and $S_3=0$ because three is odd. Thus the partial sums keep alternating between $4$ and $0$. They do not approach one finite value.

The formula can also be checked by differences. From $S_0$ to $S_1$, the change is $0-4=-4$, the second term. From $S_1$ to $S_2$, the change is $4-0=4$, the third term. The alternating terms are exactly the increments in the partial sums.

Why the Alternating Partial Sums Do Not Converge

The repeated pattern in the example is not merely a visual impression. If $a\ne0$, the even-indexed partial sums all equal $a$, while the odd-indexed partial sums all equal zero. A convergent sequence and each of its subsequences must have the same limit. These two subsequences have different limits when $a\ne0$, so the full sequence of partial sums cannot converge.

Separate the indices.
For every nonnegative integer $m$, the even-indexed partial sum is $S_{2m}=a$, and the odd-indexed partial sum is $S_{2m+1}=0$.

Assume the partial sums converge.
If $S_N$ converged to a limit $L$, then the subsequence $S_{2m}$ would also converge to $L$. But every term of that subsequence equals $a$, so its limit is $a$, and therefore $L=a$.

Use the other subsequence.
The subsequence $S_{2m+1}$ would also have to converge to $L$. Every term equals zero, so its limit is zero, and therefore $L=0$.

Conclude for nonzero first terms.
The two conclusions $L=a$ and $L=0$ are incompatible when $a\ne0$. Thus the partial sums do not converge in that case.

If $a=0$, both subsequences instead equal zero, and every term of the series is zero. Then all partial sums are zero and the series converges to zero. Hence the exact conclusion for ratio $-1$ is that the series converges if and only if its first term is zero.

Worked Example: The Zero First Term

Let $a=0$ and $r=-1$. The $k$-th term is $0\cdot(-1)^k=0$, so the series is $0+0+0+\cdots$. Its partial sum is

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

for every $N$. The even-indexed and odd-indexed partial sums both equal zero, so the partial-sum sequence has limit zero. This case matters: a ratio of negative one does not by itself force divergence if every term is zero.

How the Finite Geometric Formula Fits

For a finite geometric sum with $r\ne1$, the familiar formula is

S_N=a\frac{1-r^{N+1}}{1-r}.

The restriction $r\ne1$ is essential because the denominator $1-r$ would be zero at $r=1$. The direct calculation for ratio one is therefore $S_N=(N+1)a$, not an evaluation of this quotient at $r=1$.

At $r=-1$, the denominator is $2$, so the finite formula is valid. Substituting gives

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

If $N$ is even, then $N+1$ is odd and $(-1)^{N+1}=-1$, so $S_N=a(1-(-1))/2=a$. If $N$ is odd, then $N+1$ is even and $(-1)^{N+1}=1$, so $S_N=a(1-1)/2=0$. This reproduces the parity formula. The finite formula describes each partial sum; it does not, by itself, guarantee that the partial sums have a limit.

A Common Mistake: Confusing Alternation with Convergence

A tempting but incorrect argument is: “The terms alternate, so positive and negative contributions cancel; therefore the series sums to zero.” For $a=4$, that claim would assign the series a sum of zero. But the actual partial sums are $4,0,4,0,\ldots$. They do not approach zero or any other single number.

The correction is to use the definition of an infinite sum: inspect the limit of the partial sums. Cancellation does make every odd-indexed partial sum zero, but every even-indexed partial sum equals four. Both patterns persist indefinitely, so there is no finite series sum.

A related mistake is to substitute $r=1$ into the finite geometric quotient and treat the resulting division by zero as a sum. The quotient requires $r\ne1$. For $r=1$, count the $N+1$ constant terms directly; for $a\ne0$, the resulting partial sums $(N+1)a$ are unbounded.

Why Explicit Partial Sums Matter

These cases show why the common ratio alone is not a complete description of an infinite series. The first term also matters: when $a=0$, both unit-magnitude ratios produce only zero terms and a convergent series. For nonzero $a$, ratio one produces partial sums whose magnitudes grow without bound, while ratio negative one produces two persistent partial-sum values.

Writing out the partial sums also keeps indexing visible. With the convention used here, $S_N$ contains $N+1$ terms. That is why $S_0=a$, not an empty sum, and why the even-versus-odd rule for ratio negative one has the stated form. A different starting index can shift which partial sums take which value, but the need to examine the partial-sum sequence remains.

Takeaway: For first term $a$, a geometric series with ratio $1$ has partial sums $S_N=(N+1)a$; it converges exactly when $a=0$. With ratio $-1$, its partial sums alternate between $a$ and $0$; it also converges exactly when $a=0$.

Check Your Understanding

Use the partial-sum sequence in each question, and state whether the series has a finite sum.

For first term $a=5$ and ratio $r=1$, write $S_N$ and determine its behavior as $N$ increases.
For first term $a=-3$ and ratio $r=-1$, calculate $S_0,S_1,S_2,S_3$ and identify the two values that recur.
For first term $a=0$ and ratio $r=1$, what is every partial sum, and what is the series sum?
Why is the finite geometric quotient not applicable when $r=1$?
For $r=-1$ and $a\ne0$, which two subsequences of partial sums demonstrate nonconvergence?