What Remains After a Partial Sum?
A partial sum includes only finitely many terms of a series. When the series converges, the difference between its full sum and that partial sum is exactly the sum of the terms that were left out. For a geometric series, this omitted tail has a simple formula.
The formula is useful both for calculating a remainder and for understanding its sign and size. Careful indexing matters: after retaining \(N\) terms, the first omitted term has a factor of \(r^N\), not \(r^{N-1}\).
Here, \(N\) counts terms retained. The first term has index \(k=0\), so retaining \(N\) terms means the included indices are \(0,1,\ldots,N-1\). The first omitted term therefore has index \(k=N\). When \(N=0\), no terms have been retained and the remainder is the entire series.
The Geometric Tail Formula
For \(|r|<1\), the infinite geometric series has sum \(S=\frac{a}{1-r}\). The tail after \(N\) terms is itself geometric: it begins with \(ar^N\), and each subsequent term is obtained by multiplying by \(r\).
Because \(r\) is real and \(|r|<1\), we have \(1-r>0\). Thus the sign of \(R_N\) is determined by the sign of \(ar^N\). For a negative ratio, the remainder can be negative: in that case, the partial sum is greater than the full sum. Its size is \(|R_N|=\frac{|a||r|^N}{1-r}\).
Why the Formula Works
The tail formula follows by applying a finite geometric-sum calculation to the omitted terms, then taking a limit. Let \(M\) be a positive integer. The first \(M\) terms of the tail are \(ar^N+ar^{N+1}+\cdots+ar^{N+M-1}\). Factoring out \(ar^N\) leaves a finite geometric sum.
The sum of the first \(M\) omitted terms is \(ar^N(1+r+\cdots+r^{M-1})\).
For \(r\ne 1\), multiplying \(1+r+\cdots+r^{M-1}\) by \(1-r\) gives \(1-r^M\). Therefore the finite tail is \(\frac{ar^N(1-r^M)}{1-r}\).
Since \(|r|<1\), \(r^M\) tends to zero as \(M\) tends to infinity. The finite tail therefore tends to \(\frac{ar^N}{1-r}\), proving the formula for the infinite tail.
This limit is the tail of the series because it is the limit of the partial sums beginning at index \(N\). Also, the first \(N\) terms followed by the terms from index \(N\) onward make up the entire series. Consequently, \(S=S_N+R_N\), which gives \(R_N=S-S_N\). This proves both expressions for the remainder.
Worked Example: Finding the Tail After Four Terms
Worked Example: Finding the Tail After Four Terms
Find the remainder after the first four terms of \(12+4+\frac{4}{3}+\frac{4}{9}+\cdots\), and check it by subtracting the partial sum from the full sum.
The first term is \(a=12\), the ratio is \(r=\frac{1}{3}\), and \(N=4\). Since \(|r|=\frac{1}{3}<1\), the series converges and the tail formula applies. The first omitted term has index \(4\):
The full sum is \(S=\frac{12}{1-\frac{1}{3}}=18\). The first four terms have sum
Subtracting gives \(S-S_4=18-\frac{160}{9}=\frac{162}{9}-\frac{160}{9}=\frac{2}{9}\), agreeing with the tail formula. The first omitted term is \(\frac{4}{27}\); dividing it by \(1-\frac{1}{3}=\frac{2}{3}\) also gives \(\frac{2}{9}\).
Signed Remainders and Negative Ratios
A remainder is a difference, so it has a sign. It is not automatically a positive error. With a negative ratio, successive terms alternate in sign, and the tail's sign depends on the first omitted term. To report how far the partial sum is from the full sum without indicating direction, use the absolute error \(|S-S_N|=|R_N|\).
Worked Example: A Tail with a Negative Ratio
Find the remainder after three terms of \(5-\frac{5}{2}+\frac{5}{4}-\frac{5}{8}+\cdots\). Then check the signed difference.
Here \(a=5\), \(r=-\frac{1}{2}\), and \(N=3\). The condition holds because \(|r|=\frac{1}{2}<1\). The remainder is
The full sum is \(S=\frac{5}{1+\frac{1}{2}}=\frac{10}{3}\). The three-term partial sum is
Thus \(S-S_3=\frac{10}{3}-\frac{15}{4}=\frac{40}{12}-\frac{45}{12}=-\frac{5}{12}\). The negative sign shows that the partial sum is above the full sum. The magnitude of the error is \(\frac{5}{12}\).
Using a Tolerance to Choose the Number of Terms
The remainder formula can also answer a practical question: how many terms should be retained to make the remainder smaller than a prescribed positive tolerance? For a real geometric series with \(|r|<1\), the condition is \(\frac{|a||r|^N}{1-r}<\varepsilon\), where \(\varepsilon>0\) is the required maximum error. Check the smallest candidate \(N\) against the strict inequality, particularly when the requirement says “less than” rather than “less than or equal to.”
Worked Example: Meeting a Specified Error Tolerance
What is the smallest number of terms needed so that the error in approximating \(2+\frac{2}{3}+\frac{2}{9}+\cdots\) is less than \(\frac{1}{100}\)? Check the result and the preceding number of terms.
Here \(a=2\), \(r=\frac{1}{3}\), and \(1-r=\frac{2}{3}\). After \(N\) terms,
We need \(\frac{3}{3^N}<\frac{1}{100}\). Since both denominators are positive, this is equivalent to \(300<3^N\). For \(N=5\), \(3^5=243\), so the error is \(\frac{3}{243}=\frac{1}{81}\), which is greater than \(\frac{1}{100}\). For \(N=6\), \(3^6=729\), so the error is \(\frac{3}{729}=\frac{1}{243}\), which is less than \(\frac{1}{100}\). Since the remainder magnitude decreases as \(N\) increases when \(0<|r|<1\), six is the smallest suitable number of terms.
As a final check, the full sum is \(S=\frac{2}{1-\frac{1}{3}}=3\), and the remainder after six terms is \(R_6=\frac{1}{243}\). Therefore \(S_6=3-\frac{1}{243}=\frac{728}{243}\), and \(S-S_6=\frac{1}{243}\), as required.
A Common Indexing Mistake
The exponent in the tail formula comes from the index of the first omitted term. If the series begins with \(a\) at index zero, then \(N\) retained terms have indices \(0\) through \(N-1\). The first omitted term is \(ar^N\). Using \(ar^{N-1}\) instead counts the last included term as the first omitted term.
Worked Example: Correcting an Off-by-One Error
For \(3+\frac{3}{2}+\frac{3}{4}+\cdots\), find the tail after two terms. A common incorrect calculation uses exponent \(2-1=1\), giving \(\frac{3(\frac{1}{2})}{1-\frac{1}{2}}=3\). This is not the omitted tail: it starts with the second term, which was already included.
The correct first omitted term is \(\frac{3}{4}\), at index \(2\). Therefore
Checking by subtraction, the full sum is \(S=\frac{3}{1-\frac{1}{2}}=6\), and the first two terms sum to \(S_2=3+\frac{3}{2}=\frac{9}{2}\). Hence \(S-S_2=6-\frac{9}{2}=\frac{3}{2}\). The incorrect result \(3\) fails this check.
When the Tail Formula Applies
The condition \(|r|<1\) is essential: it guarantees that the geometric series converges and that the powers \(r^M\) in the finite-tail calculation tend to zero. Do not use the infinite-tail formula for a ratio outside this range. At the boundary \(|r|=1\), convergence must be considered separately; the formula derived here is not a justification for an infinite sum there.
Keep the signed remainder and its magnitude distinct. For instance, a negative \(R_N\) means \(S_N>S\), while \(|R_N|\) describes the distance between the two values. This distinction is important when checking an approximation or deciding whether it meets an error tolerance.
Check Your Understanding
Use the remainder as a signed difference unless a question specifically asks for its magnitude.
- For \(8+4+2+\cdots\), find the remainder after the first four terms. Check by subtracting the partial sum from the full sum.
- For \(6-2+\frac{2}{3}-\cdots\), find the signed remainder after three terms and state whether the partial sum is above or below the full sum.
- In \(a+ar+ar^2+\cdots\), which exponent gives the first omitted term after \(N\) terms, and why?
- For \(3+\frac{3}{5}+\frac{3}{25}+\cdots\), find the magnitude of the remainder after five terms.
- Why is the condition \(|r|<1\) needed to derive the infinite-tail formula?
