From a Familiar Sum to a Function
A geometric series is more than a pattern of terms: within its interval of convergence, it gives a function an exact representation as an infinite sum of powers. The basic example is the function \(1/(1-r)\). Its series has a particularly simple pattern: every coefficient is \(1\).
To understand this representation, we need to distinguish the infinite sum from its finite partial sums. A partial sum is a polynomial, so it can be calculated directly. The infinite series represents the function only where the partial sums approach the function’s value.
The Finite Geometric Sum
Begin with a finite sum, where no convergence question is involved. For a nonnegative integer \(N\), define its partial sum by
Multiplying by \(r\) shifts each power up by one. Subtracting the shifted sum cancels all the intermediate terms:
Thus, when \(r\ne 1\), factoring the left-hand side gives
This identity is an algebraic statement about a finite sum. It does not require \(|r|<1\). The restriction \(|r|<1\) becomes important when we take the limit as \(N\) increases.
Worked Example: Find a Finite Partial Sum
Find \(S_3(r)\) and verify the finite-sum formula.
By the definition of the partial sum,
The finite-sum formula gives
To check the formula, factor the numerator:
The quotient therefore agrees with the sum of its four terms whenever \(r\ne1\). At \(r=1\), the sum itself is \(1+1+1+1=4\); the quotient form is undefined, so the sum must be evaluated directly there.
Why the Infinite Series Represents the Function
For \(r\ne1\), the finite-sum identity can be rearranged to show exactly how far a partial sum is from \(1/(1-r)\):
The right-hand side is the remainder after the first \(N+1\) terms. If \(|r|<1\), then \(r^{N+1}\) approaches zero as \(N\) increases. Since \(1-r\ne0\), dividing by \(1-r\) preserves that limit. Consequently, the remainder approaches zero and the partial sums approach \(1/(1-r)\).
This proves why the representation works: the finite sum is exactly the function value minus a remainder, and the remainder vanishes for \(|r|<1\). The infinite series is therefore not merely suggested by a pattern; its value follows from a limit.
Use \(S_N(r)=1+r+\cdots+r^N\).
For \(r\ne1\), \(S_N(r)=(1-r^{N+1})/(1-r)\).
If \(|r|<1\), then \(r^{N+1}\to0\), so \(S_N(r)\to1/(1-r)\).
Center and Interval of Convergence
A power series centered at \(c\) is a series written in powers of \(r-c\). Here the terms are powers \(r^n=(r-0)^n\), so the center is \(0\). To give the interval of convergence, we must check not only the interior but also the endpoints.
When \(|r|<1\), the theorem above proves convergence. If \(|r|>1\), the terms \(r^n\) do not approach zero, so the series cannot converge. At \(r=1\), every term equals \(1\), and the partial sums are \(S_N(1)=N+1\), which grow without bound. At \(r=-1\), the partial sums alternate: \(S_0(-1)=1\), \(S_1(-1)=0\), \(S_2(-1)=1\), and \(S_3(-1)=0\). They do not approach a single limit.
The endpoint check matters even though the function \(1/(1-r)\) itself is defined at \(r=-1\), where it equals \(1/2\). The fact that a function has a value at a point does not guarantee that a particular power series converges there. At \(r=1\), the function is not defined, and the series also diverges.
Worked Example: Approximate the Function at \(r=1/3\)
Use the first four terms, from \(n=0\) through \(n=3\), to approximate \(1/(1-r)\) at \(r=1/3\). Then check the result using the exact remainder.
The partial sum is
The function value is
The remainder formula gives
Adding the remainder to the partial sum checks the function value:
Thus the four-term partial sum is \(40/27\), and its exact shortfall from the function value is \(1/54\).
Partial Sums for Negative Values
The same formula applies when \(r\) is negative. The terms then alternate in sign, but convergence still follows whenever \(|r|<1\). The remainder formula also works without changing signs by hand: its sign is determined by \(r^{N+1}\).
Worked Example: Check a Partial Sum at \(r=-2/3\)
Use \(N=3\) to find the partial sum and verify it against the function value.
First calculate each term:
The function value is
The exact remainder is
Adding this positive remainder to the partial sum gives
The check agrees exactly with the function value. Because the fourth omitted power is positive, the partial sum lies below \(3/5\).
A Common Mistake: Ignoring the Convergence Condition
A common error is to use \(1/(1-r)\) as the sum of the series for every real \(r\). For example, at \(r=2\) the function value is \(1/(1-2)=-1\). But the first partial sums are \(1\), \(1+2=3\), \(1+2+4=7\), and \(1+2+4+8=15\). They do not approach \(-1\); in fact, their terms grow.
The finite identity remains valid at \(r=2\). For example, \(S_3(2)=(1-2^4)/(1-2)=(-15)/(-1)=15\), matching the four-term sum. What fails is the step from finite sums to the infinite sum: the remainder does not tend to zero when \(|r|>1\). Always state the convergence condition before treating the geometric series as a function representation.
Why This Representation Is Useful
Each finite partial sum is a polynomial, while the infinite series represents the rational function on a specific interval. This connection lets us approximate function values using finitely many powers and quantify the difference using an exact remainder. It also gives a basic model for the larger idea of a power series: a function can be represented by infinitely many polynomial terms near its center, provided the series converges there.
For this series, the terms and the remainder are especially easy to compute. The essential habits are broadly useful: identify the center, establish the interval of convergence, check the endpoints, and keep the finite-sum identity separate from the infinite-series conclusion.
Check Your Understanding
Answer each question using the finite-sum formula and the convergence condition where appropriate.
- What are the center and interval of convergence of \(\sum_{n=0}^{\infty}r^n\)?
- Write \(S_4(r)\) as a sum and as a quotient.
- Why does \(\sum_{n=0}^{\infty}r^n=1/(1-r)\) require \(|r|<1\)?
- Does the series converge at \(r=-1\), even though \(1/(1-r)\) is defined there? Explain using partial sums.
- For \(r=1/2\), write the exact remainder after the partial sum \(S_2(r)\).
