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Calculus BC · Unit 10 · 10.10
AP Calculus BC · Infinite Sequences and Series
The Alternating Series Error Bound
Truncating costs less than the very next term — and its sign tells you which way you erred.
key idea$\big\lvert S-S_N\big\rvert\le b_{N+1}\quad\text{— the first term you left out}$
The rule
For a convergent alternating series, truncating after $N$ terms costs you less than the very next term. Nothing needs to be computed beyond writing that term down — which makes this the easiest error bound in the course, and the one most often misindexed.
| Symbol | Meaning | Note |
|---|
| $S$ | the true sum | usually unknown |
| $S_N$ | the sum of the first $N$ terms | what you actually computed |
| $b_{N+1}$ | the absolute value of the next term | the bound |
| $\lvert S-S_N\rvert$ | the truncation error | $\le b_{N+1}$ |
| the sign of $S-S_N$ | matches the first omitted term | so $S$ lies between $S_N$ and $S_{N+1}$ |
Row 5 is the extra information most students leave on the table: not only is the error small, but you know which side of the true sum your estimate falls on.
four facts
- The bound is the first omitted term, in absolute value. Count carefully — off-by-one here is the standard error.
- The two alternating-series conditions must hold; the bound is meaningless without them.
- To guarantee an error under $\varepsilon$, solve $b_{N+1}\lt\varepsilon$ for $N$.
- The true sum always lies between consecutive partial sums, which brackets it from both sides.
Truncation errorThe gap between the true sum and a partial sum. It is what this bound controls.
$b_{N+1}$The magnitude of the first term not included in $S_N$.
BracketingConsecutive partial sums lie on opposite sides of $S$, so any two of them trap it.
SharpnessThe bound is usually generous — the actual error is smaller, often much smaller.
Pattern recognition
| Question | What to solve |
|---|
| “bound the error in $S_4$” | compute $b_5$ |
| “how many terms for error $\lt0.01$?” | solve $b_{N+1}\lt0.01$ |
| “is $S_N$ an over- or underestimate?” | read the sign of the first omitted term |
| “between which two values does $S$ lie?” | between $S_N$ and $S_{N+1}$ |
All four are answered by looking at one term: the first one you did not include. Identifying it correctly is the entire skill.
Worked examples
EX 1
Bound the error of a partial sum
the next termEstimate $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}$ using the first four terms, and bound the error.
1
Compute $S_4$
$1-\dfrac12+\dfrac13-\dfrac14=\dfrac{7}{12}\approx0.5833$.
2
Identify the first omitted term
The fifth term, $\dfrac{(-1)^{6}}{5}=+\dfrac15$.
3
State the bound
$\lvert S-S_4\rvert\le\dfrac15=0.2$.
4
Determine the direction
The omitted term is positive, so $S_4$ is an underestimate.
5
Bracket the sum
$0.5833\lt S\lt0.7833$. The true value is $\ln2\approx0.6931$. ✓
EX 2
How many terms are needed?
solve for $N$How many terms of $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^{2}}$ guarantee an error below $0.01$?
1
Write the bound
$b_{N+1}=\dfrac{1}{(N+1)^{2}}$.
2
Set the inequality
$\dfrac{1}{(N+1)^{2}}\lt0.01$.
3
Solve
$(N+1)^{2}\gt100$, so $N+1\gt10$ and $N\gt9$.
4
Choose the smallest integer
$N=10$.
5
State the answer carefully
Ten terms suffice. Answering $11$ confuses the index of the omitted term with the count of terms used.
EX 3
Over or under?
the sign of the omitted termFor $\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{2^{n}}$, is $S_3$ an overestimate or an underestimate?
1
Write out the terms
$1,\ -\dfrac12,\ \dfrac14,\ -\dfrac18,\ \dfrac{1}{16},\dots$
2
Compute $S_3$
The first three terms: $1-\dfrac12+\dfrac14=\dfrac34$.
3
Identify the first omitted term
$-\dfrac18$ — the fourth term.
4
Read the sign
Negative, so adding it would decrease the estimate. $S_3$ is an overestimate.
5
Check against the exact sum
This is geometric with $r=-\dfrac12$, so $S=\dfrac{1}{1+\tfrac12}=\dfrac23\approx0.667\lt0.75$. ✓
EX 4
When the bound does not apply
the conditions matterCan the alternating series error bound be used for $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\dfrac{n}{n+1}$?
1
Check the limit condition
$b_n=\dfrac{n}{n+1}\to1\neq0$.
2
Note the consequence
The alternating series test fails, and the series in fact diverges.
3
Draw the conclusion
There is no sum $S$ to approximate, so no error to bound.
4
State the requirement
The bound applies only to series that satisfy the alternating series test.
5
Note the practical check
Confirm $b_n$ decreases to $0$ before quoting any error bound. Quoting it for a divergent series is a meaningless statement.
Watch out
The bound is the term you left out
For $S_4$ the bound is $b_5$, not $b_4$. Writing out the first several terms and pointing at the next one prevents the off-by-one.
The bound requires the test's conditions
$b_n$ must decrease to zero. Without that, the estimate is not guaranteed — and there may be no sum at all.
The sign tells you the direction
A positive first omitted term makes $S_N$ an underestimate; a negative one makes it an overestimate. This is worth stating on free response.
The bound is generous
The actual error is typically well under $b_{N+1}$. That is fine — a guarantee is what is being asked for, not the exact error.