The Geometric Series and Its Remainder

From a Finite Sum to an Infinite Series

A geometric sum is a sum in which each term is obtained from the preceding term by multiplying by the same number. For example, \(1+u+u^2+u^3\) is a finite geometric sum with ratio \(u\). Its finite version has a simple exact formula, and that formula reveals when an infinite geometric series can represent a function.

The essential issue is not just finding a formula for a finite sum. It is determining what happens to the part left out when the sum is truncated. That remainder both justifies the infinite-series identity and measures the error in a numerical approximation.

The Finite Geometric-Sum Identity

For \(u\ne 1\), the finite sum is given by the following identity:

The identity follows from cancellation. Multiplying the sum by \(1-u\) gives one copy of the sum minus a shifted copy. All intermediate powers cancel.

The condition \(u\ne1\) matters: it is required to divide by \(1-u\). When \(u=1\), the sum is instead \(S_N=N+1\), since every one of its \(N+1\) terms equals \(1\).

The Exact Remainder

To approximate the infinite series, it is useful to know exactly what is omitted after the partial sum \(S_N\). For \(u\ne1\), the finite identity gives

This expression is the exact remainder, or truncation error, when \(S_N\) is used to approximate \(1/(1-u)\). It is signed: it records whether the partial sum lies above or below the value as well as the size of the difference.

When \(|u|<1\), the triangle inequality gives \(|1-u|\ge 1-|u|>0\). Therefore the exact error also satisfies a convenient bound that depends only on \(|u|\):

The bound is valid for \(|u|<1\). For a particular \(u\), the exact error formula may give a sharper value. The bound is especially useful when we want a guaranteed error estimate without calculating the signed remainder.

When the Infinite Geometric Series Is Valid

An infinite series is defined through the limit of its partial sums. Thus, to identify the infinite geometric series with \(1/(1-u)\), we must show that the partial sums approach that value. The exact remainder makes the needed condition visible: the remainder tends to zero when \(|u|<1\).

Here is a justification that includes why the powers vanish. If \(u=0\), then \(u^{N+1}=0\) for every \(N\). Otherwise let \(q=|u|\), so \(0<q<1\), and set \(a=1/q-1>0\). Then \(q=1/(1+a)\). The binomial inequality \((1+a)^m\ge 1+ma\) for positive integers \(m\) implies

The right-hand side tends to zero as \(m\) increases, so \(q^m\), and hence \(|u^m|\), tends to zero. Applying this with \(m=N+1\), the exact remainder \(R_N=u^{N+1}/(1-u)\) tends to zero because \(1-u\) is a fixed nonzero number. Since \(S_N=1/(1-u)-R_N\), the partial sums tend to \(1/(1-u)\). This proves the theorem.

Why the Condition on the Ratio Matters

The finite-sum formula holds for every \(u\ne1\), but the infinite-series conclusion does not. It is the vanishing of \(u^{N+1}\), not merely the existence of a finite formula, that allows the remainder to disappear. If \(|u|\ge1\), the powers do not tend to zero in general, so the argument cannot establish the infinite identity.

For example, with \(u=2\), the finite formula gives \(S_N=(1-2^{N+1})/(1-2)=2^{N+1}-1\). These partial sums grow rather than approach \(-1=1/(1-2)\). The exact difference from \(-1\) is \(R_N=2^{N+1}/(1-2)=-2^{N+1}\), which does not tend to zero.

Using Remainders in Practice

The remainder formula is useful in two complementary ways. If \(u\) and \(N\) are known, it gives the exact difference between the target value and the partial sum. If only a guaranteed tolerance is needed, the bound \(|u|^{N+1}/(1-|u|)\) provides a sufficient test. A smaller \(|u|\) makes the powers shrink faster; for a ratio close to \(1\) in absolute value, more terms may be required.

Be careful with the indexing. A partial sum from \(k=0\) through \(k=N\) contains \(N+1\) terms, and its first omitted power is \(u^{N+1}\). Consequently, the remainder uses the exponent \(N+1\), not \(N\). Keeping the endpoint and the first omitted term visible helps prevent off-by-one errors.

The central chain of reasoning is: establish the finite identity, retain the exact remainder, and then verify that the remainder tends to zero under the condition \(|u|<1\). The same remainder also supplies a quantitative error estimate for finite approximations.