Why Compare Finite and Infinite Sums?
A finite geometric sum contains a fixed number of terms. An infinite geometric series contains every term in the pattern. When the terms shrink toward zero quickly enough, the infinite sum can provide a convenient approximation to a long finite sum. But the two sums are not automatically equal: the terms left out of the finite sum make up a remainder.
Tracking that remainder makes the comparison precise. It shows not only the value of the infinite sum, but also how far it is from a chosen finite partial sum and when that difference is small enough to ignore.
Finite Sums and Their Infinite Counterparts
Let \(a\) be the first term and \(r\) the common ratio. The first \(n\) terms, starting with the power \(r^0\), form the finite sum
Here \(n\) is a positive integer. The corresponding infinite series continues with \(ar^n, ar^{n+1},\ldots\). When \(|r|<1\), it converges to \(S_\infty=\frac{a}{1-r}\). For \(r\ne 1\), the finite sum is \(S_n=\frac{a(1-r^n)}{1-r}\). The finite formula works whether or not the corresponding infinite series converges; if \(r=1\), instead \(S_n=na\).
The word “remainder” here has a specific meaning: it records the infinite sum minus the finite sum, in that order. For a negative ratio, the omitted terms may begin by decreasing the finite sum rather than increasing it, so the sign matters.
Deriving the Remainder Formula
Assume \(|r|<1\). The terms omitted after the first \(n\) terms are \(ar^n+ar^{n+1}+ar^{n+2}+\cdots\). Factoring out their first term leaves another geometric series, with first term \(ar^n\) and common ratio \(r\). Since \(|r|<1\), its sum is \(ar^n/(1-r)\).
This calculation also proves that the omitted-term interpretation and the difference of the two sums agree. Subtracting the finite sum from the infinite sum cancels the first \(n\) terms, leaving exactly the terms starting at \(ar^n\).
The condition \(|r|<1\) is essential for the infinite sum and remainder formulas: it ensures that the infinite tail converges. The finite formula has a wider scope. For \(r\ne1\), it applies for any real \(r\), including values for which the associated infinite series does not converge.
Worked Examples
Worked Example: Finding the Omitted Tail
Compare the first four terms of \(6+2+\frac23+\frac29+\cdots\) with the infinite sum. Here \(a=6\), \(r=\frac13\), and \(n=4\). Because \(|r|=\frac13<1\), the infinite series converges.
First calculate the finite sum directly:
The infinite sum is
Now use the remainder formula. The first omitted term has power \(r^4\), since four terms use powers \(r^0\) through \(r^3\):
Check by subtraction: \(S_\infty-S_4=9-\frac{80}{9}=\frac{81}{9}-\frac{80}{9}=\frac19\). Thus the infinite sum exceeds the four-term sum by \(\frac19\).
Worked Example: A Negative Ratio and a Signed Remainder
Compare the first three terms of \(5-\frac52+\frac54-\frac58+\cdots\) with its infinite sum. Here \(a=5\), \(r=-\frac12\), and \(n=3\). Since \(|r|=\frac12<1\), the infinite sum exists.
The three-term sum is
The infinite sum is
The signed remainder is
Check the difference: \(S_\infty-S_3=\frac{10}{3}-\frac{15}{4}=\frac{40}{12}-\frac{45}{12}=-\frac{5}{12}\), as required. The negative sign means the infinite sum is less than the three-term sum. The absolute error is \(|R_3|=\frac{5}{12}\).
Worked Example: Checking an Approximation Tolerance
How many terms of \(4+2+1+\frac12+\cdots\) are needed so that the infinite sum differs from the finite sum by less than \(\frac{1}{100}\)? Here \(a=4\), \(r=\frac12\), and the infinite sum is \(4/(1-\frac12)=8\).
After \(n\) terms, the absolute error is
We require \(8(\frac12)^n<\frac{1}{100}\). Check consecutive integer values. For \(n=9\), the error is \(8/512=1/64\), and \(\frac{1}{64}>\frac{1}{100}\). For \(n=10\), the error is \(8/1024=1/128\), and \(\frac{1}{128}<\frac{1}{100}\). Since \(8(\frac12)^n\) decreases as \(n\) increases, ten terms suffice and nine do not.
Check the ten-term partial sum against the remainder: \(S_{10}=8(1-(\frac12)^{10})=8-\frac{1}{128}=\frac{1023}{128}\). Therefore \(8-S_{10}=\frac{1024}{128}-\frac{1023}{128}=\frac{1}{128}\), which is below the requested tolerance.
When Is the Infinite Sum a Valid Replacement?
For \(|r|<1\), using \(S_\infty\) in place of \(S_n\) is an approximation whose exact signed difference is \(R_n\). It is appropriate when the size of that difference meets the accuracy required for the problem. For a tolerance \(\varepsilon>0\), the precise condition is
If the question asks for the exact sum of the first \(n\) terms, replacing it with the infinite sum is not generally valid: the omitted tail must be zero for the two values to agree. If \(a=0\), every term is zero. If \(r=0\) and \(n\ge1\), all terms after the first are zero, so the finite and infinite sums also agree. Otherwise, when \(|r|<1\), the tail is generally nonzero, even if it is very small.
If \(|r|\ge1\), do not use the convergent infinite-sum formula as a substitute. Except for the zero sequence \(a=0\), the geometric series does not converge in these cases. A finite sum still has a well-defined value, but there is no finite infinite sum to replace it with.
Common Pitfall: Starting the Remainder at the Wrong Power
After \(n\) terms starting with \(ar^0\), the last included term is \(ar^{n-1}\), and the first omitted term is \(ar^n\). A common indexing error is to use \(ar^{n-1}\) as the first omitted term. For example, after four terms of \(6+2+\frac23+\frac29+\cdots\), that error would start the tail with \(\frac29\), which is already included in the finite sum.
The correct omitted tail starts with \(6(\frac13)^4=\frac{2}{27}\), and its sum is \(\frac{2}{27}/(1-\frac13)=\frac19\). The check \(S_\infty-S_4=9-\frac{80}{9}=\frac19\) confirms the correct starting power.
Summary
A finite geometric sum includes a specified number of terms; its infinite counterpart includes the tail as well. When \(|r|<1\), the difference is exactly \(\frac{ar^n}{1-r}\) after \(n\) terms, and the absolute error is \(\frac{|a||r|^n}{|1-r|}\). The finite sum and infinite sum are equal only when the omitted tail is zero. Otherwise, replacing one with the other is justified only when the resulting error is acceptable.
Check Your Understanding
Use the remainder to compare each finite sum with its infinite counterpart where the series converges.
- For \(3+\frac32+\frac34+\cdots\), write the sum of the first five terms and identify the first omitted term.
- For a geometric series with \(a=2\), \(r=-\frac13\), and \(n=4\), find the signed remainder and the absolute error.
- Why does the remainder after \(n\) terms begin with \(ar^n\), rather than \(ar^{n-1}\)?
- State the condition on \(r\) that permits use of the infinite geometric-sum formula, and explain what that condition guarantees.
- What inequality must hold for the infinite sum to approximate a finite sum within a specified tolerance \(\varepsilon>0\)?
