Geometric Series

A Series Built from a Constant Ratio

In the previous tutorial, the Necessary Condition for Series Convergence showed that a convergent series must have terms tending to zero. Geometric series provide a useful setting in which to see how the behavior of the terms controls the behavior of the partial sums. Each term is obtained from the preceding term by multiplication by one fixed real number.

Let $a$ be the initial term and $r$ the fixed ratio. We write the geometric series as $\sum_{n=0}^{\infty} ar^n$, with the convention $r^0=1$. When $r=0$, this means the terms are $a,0,0,\ldots$. More generally, the terms can be specified without using powers: set $t_0=a$, and define each next term by $t_{n+1}=rt_n$. This recurrence is the essential property.

Definition: A geometric series is a series whose terms satisfy $t_0=a$ and $t_{n+1}=rt_n$ for a fixed $r\in\mathbb{R}$. It is conventionally written $\sum_{n=0}^{\infty} ar^n$, where $a$ is the initial term and $r$ is the ratio parameter.

The recurrence $t_{n+1}=rt_n$ remains meaningful even when a term is zero. It is safer than describing $r$ as the quotient of consecutive terms: such a quotient is defined only when the preceding term is nonzero. For instance, if $a\ne0$ and $r=0$, the terms after the first are all zero, and the quotient of one such term by the preceding zero term is undefined.

For $N\geq0$, let $S_N=\sum_{n=0}^{N} ar^n$ be the $N$th partial sum. Convergence of the series means that the sequence of these partial sums has a finite limit. The first step toward understanding that sequence is an identity for finite sums.

The Finite Geometric-Sum Identity

Proposition (Finite Geometric-Sum Identity): Let $k\geq1$ be an integer. If $r\ne1$, then $$ \sum_{j=0}^{k-1}r^j=\frac{1-r^k}{1-r}. $$ If $r=1$, then $\sum_{j=0}^{k-1}r^j=k$.

Proof. Write $T=\sum_{j=0}^{k-1}r^j=1+r+\cdots+r^{k-1}$. Multiplying by $r$ gives $rT=r+r^2+\cdots+r^k$. Subtracting the second expression from the first cancels all the intermediate terms:

T-rT=1-r^k, \qquad (1-r)T=1-r^k.

When $r\ne1$, division by $1-r$ gives the stated identity. When $r=1$, each of the $k$ terms in the sum equals $1$, so the sum is $k$. This separate case matters: the displayed fraction is not defined at $r=1$. $\square$

Multiplying the identity by $a$ gives the partial sums of a geometric series. In particular, for $r\ne1$,

S_N=\sum_{n=0}^{N}ar^n =a\frac{1-r^{N+1}}{1-r}.

There are $N+1$ terms here, so the exponent in the numerator is $N+1$. If $r=1$, each term is $a$, and instead $S_N=(N+1)a$. These formulas describe finite sums; the convergence question asks what happens as $N$ increases without bound.

Worked Example: A Positive Ratio Below One

Consider the series $\sum_{n=0}^{\infty}5(1/3)^n$. Its first five terms, from $n=0$ through $n=4$, are $5,5/3,5/9,5/27,5/81$. The finite identity gives

S_4 =5\frac{1-(1/3)^5}{1-1/3} =5\frac{242/243}{2/3} =\frac{605}{81}.

Checking by a common denominator gives the same result: $5+5/3+5/9+5/27+5/81=(405+135+45+15+5)/81=605/81$. Since the ratio has absolute value less than one, the convergence theorem below shows that the sequence of partial sums has a finite limit.

When Does a Geometric Series Converge?

The key distinction is whether successive powers of $r$ shrink toward zero. If $|r|<1$, then $|r|^n$ tends to zero. If $|r|\geq1$, the terms of a nonzero geometric series do not tend to zero. The convergence criterion makes this distinction precise.

Theorem (Convergence Criterion for a Geometric Series): If $a=0$, the geometric series $\sum_{n=0}^{\infty} ar^n$ converges for every $r\in\mathbb{R}$. If $a\ne0$, it converges if and only if $|r|<1$.

Proof. If $a=0$, every term is zero, so every partial sum is zero and the series converges. Now suppose $a\ne0$.

First let $|r|<1$. We verify the Cauchy Criterion for Series. Choose integers $M>N\geq0$. The sum from the $(N+1)$st term through the $M$th term is a finite geometric sum:

\left|\sum_{n=N+1}^{M}ar^n\right| =|a|\,|r|^{N+1} \left|\sum_{j=0}^{M-N-1}r^j\right|.

There are $M-N\geq1$ terms in the sum over $j$. Since $|r|<1$, we have $r\ne1$, so the finite identity applies. For $k=M-N$, the triangle inequality gives

\left|\sum_{j=0}^{k-1}r^j\right| =\left|\frac{1-r^k}{1-r}\right| \leq\frac{1+|r|^k}{|1-r|} \leq\frac{2}{|1-r|}.

Consequently, for all $M>N\geq0$,

\left|\sum_{n=N+1}^{M}ar^n\right| \leq \frac{2|a|}{|1-r|}|r|^{N+1}.

Because $|r|<1$, the powers $|r|^{N+1}$ tend to zero. Given $\varepsilon>0$, choose $N_0$ sufficiently large that $\frac{2|a|}{|1-r|}|r|^{N_0+1}<\varepsilon$. Then for every $M>N\geq N_0$, the block of terms from $N+1$ through $M$ has absolute value less than $\varepsilon$. The Cauchy Criterion for Series therefore implies convergence.

For the converse, suppose $|r|\geq1$. If $r=1$, every term equals $a\ne0$. If $r=-1$, the terms alternate between $a$ and $-a$, so they do not tend to zero. If $|r|>1$, then $|ar^n|=|a||r|^n\geq|a|>0$ for every $n$, so again the terms do not tend to zero. In every case the Necessary Condition for Series Convergence fails. Thus the series diverges whenever $|r|\geq1$, proving the criterion. $\square$

The proof in the convergent case used finite sums to control every sufficiently late block of terms, not just a single term. This is exactly the kind of estimate required by the Cauchy Criterion. It also handles $r=0$: once $N\geq0$, the block beginning at $N+1$ consists entirely of zeros, and the displayed bound remains valid.

Worked Example: A Negative Ratio with Absolute Value Below One

Consider $\sum_{n=0}^{\infty}4(-1/2)^n$. Its terms begin $4,-2,1,-1/2,1/4$. Their signs alternate, but the ratio satisfies $|-1/2|=1/2<1$, so the convergence criterion proves that the series converges. For example, the partial sum through $n=3$ is

S_3=4-2+1-\frac12=\frac52.

The finite identity verifies this calculation: $S_3=4(1-(-1/2)^4)/(1-(-1/2))=4(15/16)/(3/2)=5/2$. Alternating signs alone do not determine convergence; here it is the shrinking absolute values of the terms, captured by $|r|<1$, that yield convergence.

Worked Example: Zero as the Ratio

Take $a=7$ and $r=0$. The recurrence gives $t_0=7$, $t_1=0\cdot7=0$, and $t_n=0$ for every $n\geq1$, because $0\cdot0=0$. Thus the partial sums satisfy $S_0=7$ and $S_N=7$ for every $N\geq1$. The series converges. No quotient of consecutive terms is needed to describe it; after the first term, such quotients would involve division by zero.

Worked Example: Ratios on the Boundary

Let $a=2$. When $r=1$, every term is $2$, and $S_N=2(N+1)$, which grows without bound. When $r=-1$, the terms are $2,-2,2,-2,\ldots$; the terms fail to tend to zero, and the partial sums alternate between $2$ and $0$. In both cases $|r|=1$, so the convergence criterion says the series diverges. The two boundary cases behave differently as partial-sum sequences, but neither series converges.

How to Use the Criterion

For a nonzero initial term, test the absolute value of the ratio—not merely whether the ratio itself is less than one. A negative ratio such as $-1/2$ has absolute value less than one and gives a convergent series. A ratio such as $-2$ is less than one as a real number, but its absolute value is greater than one, so the corresponding nonzero geometric series diverges.

The case $a=0$ is a genuine exception to the “if and only if” condition involving $|r|$. If the initial term is zero, the recurrence makes every term zero, whatever the value of $r$. This is why the theorem separates $a=0$ from $a\ne0$.

The finite-sum identity is also useful beyond convergence. It turns a block of geometric terms into an expression involving powers of $r$, which can then be estimated. In the next tutorial, the limiting behavior of the partial-sum formula will be examined directly to establish the value of a convergent geometric series.

Takeaway: A geometric series is governed by the recurrence $t_{n+1}=rt_n$. If its initial term is nonzero, the series converges exactly when $|r|<1$; if its initial term is zero, it converges trivially for every ratio. The finite-sum identity supplies the block estimates needed for the convergence proof.

Check Your Understanding

Use the finite-sum identity and the convergence criterion to answer the following questions.

Why is the recurrence $t_{n+1}=rt_n$ preferable to defining the ratio as a quotient of consecutive terms when $r=0$?
In the finite geometric-sum identity, why must $r=1$ be treated separately?
For a geometric series with nonzero initial term and ratio $-3/4$, does the series converge? Which quantity determines the answer?
Explain how the Cauchy Criterion is used to prove convergence when $|r|<1$.
Why does the Necessary Condition for Series Convergence rule out convergence when $a\ne0$ and $|r|\geq1$?

What You'll Learn