The Central Idea
A positive-term series accumulates in one direction. An alternating series repeatedly adds and subtracts.
\[ a_1-a_2+a_3-a_4+\cdots \]If the term magnitudes decrease toward zero, the positive and negative contributions can trap the partial sums in a shrinking interval.
What Is an Alternating Series?
An alternating series is a series whose nonzero terms switch sign from positive to negative or from negative to positive.
A common form is
\[ \sum_{n=1}^{\infty}(-1)^{n-1}b_n, \]where \(b_n\ge 0\).
Two Standard Sign Patterns
The pattern
\[ (-1)^{n-1} \]begins positive:
\[ 1,-1,1,-1,\ldots \]The pattern
\[ (-1)^n \]begins negative:
\[ -1,1,-1,1,\ldots \]The Magnitude Sequence
When a series is written as
\[ \sum (-1)^{n-1}b_n, \]the sequence \(b_n\) describes the magnitudes of the alternating terms.
The Alternating Series Test is stated in terms of this positive sequence.
A First Example
Consider
\[ 1-\frac12+\frac13-\frac14+\frac15-\cdots. \]This is the alternating harmonic series:
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}. \]Here,
\[ b_n=\frac1n. \]The Terms Must Still Approach Zero
Alternating signs do not override the basic necessary condition for series convergence.
If
\[ \sum a_n \]converges, then
\[ \boxed{\lim_{n\to\infty}a_n=0}. \]Alternation Alone Is Not Enough
Consider
\[ 1-1+1-1+\cdots. \]The terms alternate, but their magnitudes remain \(1\).
Since the terms do not approach zero, the series diverges.
The Alternating Series Test
Suppose
\[ \sum_{n=1}^{\infty}(-1)^{n-1}b_n \]has \(b_n\ge0\). If
- \(b_{n+1}\le b_n\) eventually, and
- \(\displaystyle \lim_{n\to\infty}b_n=0\),
then the series converges.
The Test Has Two Main Conditions
The magnitude sequence must:
- decrease eventually;
- approach zero.
Both conditions matter.
Why “Eventually Decreasing” Is Enough
A finite number of unusual early terms does not determine convergence.
If \(b_n\) becomes decreasing after some index \(N\), that is sufficient for the Alternating Series Test.
Finite Changes Do Not Affect Convergence
Adding, removing, or changing finitely many terms changes the value of a convergent series but does not change whether the infinite tail converges.
This is why eventual behavior is the central issue.
Example: Alternating Harmonic Series
Consider
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}. \]The magnitude sequence is
\[ b_n=\frac1n. \]It decreases and approaches zero. Therefore, the series converges.
A Contrast with the Harmonic Series
The ordinary harmonic series
\[ \sum_{n=1}^{\infty}\frac1n \]diverges.
But the alternating version
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n \]converges because cancellation changes the behavior of the partial sums.
Partial Sums Reveal the Mechanism
For the alternating harmonic series,
\[ S_1=1, \] \[ S_2=1-\frac12, \] \[ S_3=1-\frac12+\frac13, \] \[ S_4=1-\frac12+\frac13-\frac14. \]The odd and even partial sums approach the same limiting value from opposite sides.
Odd Partial Sums
Under the Alternating Series Test conditions, the odd partial sums form one monotone sequence.
In a series that begins positive, they typically move downward toward the sum.
Even Partial Sums
The even partial sums form another monotone sequence.
In a series that begins positive, they typically move upward toward the same sum.
The Sum Is Trapped
When the magnitude sequence decreases, the sum lies between consecutive partial sums:
\[ S_{2k}\le S\le S_{2k+1} \]for a standard alternating series beginning with a positive term.
Why the Trap Shrinks
The distance between consecutive partial sums is the magnitude of the next term:
\[ |S_{n+1}-S_n|=b_{n+1}. \]Since \(b_{n+1}\to0\), the trapping interval becomes arbitrarily small.
Example: A Rational Magnitude
Consider
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+1}. \]The magnitude is
\[ b_n=\frac{n}{n^2+1}. \]Since \(b_n\to0\) and is eventually decreasing, the series converges by the Alternating Series Test.
Checking the Limit Condition
For
\[ b_n=\frac{n}{n^2+1}, \]divide numerator and denominator by \(n^2\):
\[ b_n=\frac{1/n}{1+1/n^2}\to0. \]Checking Decrease with Calculus
When a formula is cumbersome, define a continuous function \(f(x)\) with \(f(n)=b_n\).
If
\[ f'(x)<0 \]for sufficiently large \(x\), then \(b_n\) is eventually decreasing.
Example of Derivative-Based Verification
Let
\[ f(x)=\frac{x}{x^2+1}. \]Then
\[ f'(x)=\frac{1-x^2}{(x^2+1)^2}. \]For \(x>1\), \(f'(x)<0\), so the sequence is decreasing eventually.
Checking Decrease Without Calculus
Sometimes it is easier to compare consecutive terms directly.
To show
\[ b_{n+1}\le b_n, \]substitute the formulas and simplify.
Example: Consecutive-Term Comparison
For
\[ b_n=\frac1{\sqrt n}, \]we have
\[ \sqrt{n+1}>\sqrt n, \]so
\[ \frac1{\sqrt{n+1}}<\frac1{\sqrt n}. \]Example: Alternating \(p\)-Type Series
Consider
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^p}. \]If \(p>0\), then \(1/n^p\) decreases to zero.
Therefore the series converges by the Alternating Series Test.
A Useful Threshold
For
\[ \sum\frac{(-1)^{n-1}}{n^p}, \]the alternating series converges for every
\[ p>0. \]Later we will distinguish what happens when the corresponding positive series is also convergent.
Example: Divergence Because the Limit Fails
Consider
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n+1}. \]Since
\[ \frac{n}{n+1}\to1, \]the terms do not approach zero. The series diverges.
Do Not Apply the Alternating Series Test When the Limit Is Nonzero
If
\[ \lim b_n=L\ne0, \]the series fails the necessary term condition immediately.
No further convergence test is needed.
What If the Magnitudes Are Not Decreasing?
Failure of the monotonicity condition means only that the Alternating Series Test cannot be used directly.
It does not automatically prove divergence.
Inconclusive Is Not the Same as Divergent
A convergence test can have three practical outcomes:
- it proves convergence;
- it proves divergence;
- its hypotheses fail, so it is inconclusive.
The Alternating Series Test mainly proves convergence.
Alternating Series Estimation Theorem
Suppose an alternating series satisfies the Alternating Series Test and
\[ S_N \]is the \(N\)-th partial sum. Then the remainder
\[ R_N=S-S_N \]satisfies
\[ \boxed{|R_N|\le b_{N+1}}. \]The First Omitted Term Controls the Error
For a convergent alternating series with decreasing term magnitudes, the magnitude of the error after \(N\) terms is no larger than the magnitude of the next term.
This makes alternating series especially convenient for approximation.
Example: Error Bound
Suppose
\[ S=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}. \]If we stop after \(10\) terms, then
\[ |R_{10}|\le\frac1{11}. \]Choosing the Number of Terms
To guarantee
\[ |R_N|<\varepsilon, \]it is enough to choose \(N\) so that
\[ b_{N+1}<\varepsilon. \]Example: Required Accuracy
For
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}, \]suppose we want error less than \(0.01\).
Require
\[ \frac1{N+1}<0.01. \]Thus \(N+1>100\), so \(N=100\) terms are sufficient.
The Error Also Has a Predictable Sign
In a standard alternating series with decreasing magnitudes, the true sum lies between consecutive partial sums.
Therefore the next omitted term points in the direction from \(S_N\) toward the true sum.
Odd and Even Approximations
If the series begins positive, one family of partial sums lies above the true sum and the other lies below it.
This gives both an approximation and an interval containing the exact sum.
A Two-Sided Bound
If \(S_N\) and \(S_{N+1}\) are consecutive partial sums of a convergent alternating series with decreasing magnitudes, then
\[ \min(S_N,S_{N+1})\le S\le\max(S_N,S_{N+1}). \]Example: Bounding an Unknown Sum
Suppose
\[ S_4=0.5833 \]and
\[ S_5=0.7833 \]are consecutive partial sums satisfying the alternating-series conditions. Then
\[ 0.5833\le S\le0.7833. \]Alternating Structure Can Be Hidden
Some series are alternating even if the factor \((-1)^n\) is not written explicitly.
For example,
\[ -\frac12+\frac13-\frac14+\frac15-\cdots \]can be rewritten with an alternating sign factor before applying the test.
Reindexing an Alternating Series
A shift in the starting index may change whether the sign factor is written as \((-1)^n\) or \((-1)^{n-1}\), but it does not change the underlying convergence behavior.
Starting Positive or Negative Does Not Matter for Convergence
Multiplying an entire series by \(-1\) changes its sum by a factor of \(-1\), but it does not change whether the series converges.
Connection to Positive-Term Tests
Before relying only on alternating behavior, it can be useful to inspect
\[ \sum |a_n|. \]If the corresponding positive-term series already converges, then the alternating series certainly converges.
The next lesson will formalize this distinction.
A Preview of Absolute Convergence
A series
\[ \sum a_n \]is called absolutely convergent when
\[ \sum |a_n| \]converges.
We will study this systematically in the next lesson.
A Preview of Conditional Convergence
Some alternating series converge even though the corresponding series of absolute values diverges.
The alternating harmonic series is the standard example.
Example: A Rapidly Decreasing Alternating Series
Consider
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac1{n^3}. \]The magnitude \(1/n^3\) decreases to zero, so the Alternating Series Test proves convergence.
Example: A Slowly Decreasing Alternating Series
Consider
\[ \sum_{n=2}^{\infty}\frac{(-1)^n}{\ln n}. \]Since \(1/\ln n\to0\) and decreases for \(n>1\), the series converges by the Alternating Series Test.
Example: Alternating Radical Terms
Consider
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n+2}}. \]The magnitudes decrease and approach zero, so the series converges.
Example: Alternating Rational Terms
Consider
\[ \sum_{n=1}^{\infty}(-1)^n\frac{2n+1}{n^2+4}. \]The magnitude behaves like
\[ \frac{2n}{n^2}=\frac2n, \]so it approaches zero and is eventually decreasing. The alternating series converges.
Recognizing Eventual Decrease from Dominant Terms
For many rational expressions,
\[ b_n=\frac{P(n)}{Q(n)}, \]dominant powers give a quick first picture of large-\(n\) behavior.
A formal monotonicity check may still be needed, but asymptotic structure often predicts the result.
Oscillation and Convergence Are Different Ideas
The terms of an alternating series oscillate in sign.
The partial sums can still converge because the size of each correction becomes smaller.
Why Large Early Terms Do Not Matter
A convergent alternating tail can be preceded by any finite collection of terms.
Those early terms shift the total sum but do not change convergence.
Alternation Need Not Begin at \(n=1\)
A series may become alternating only after some finite starting region.
If its tail satisfies the Alternating Series Test, then the whole series converges.
A Reliable Procedure
- Identify the sign pattern.
- Write the term magnitude as \(b_n\ge0\).
- Check whether \(b_n\to0\).
- Check whether \(b_n\) is eventually decreasing.
- Conclude convergence if both conditions hold.
- If approximation is required, use \(|R_N|\le b_{N+1}\).
- If a hypothesis fails, do not automatically claim divergence.
A Fast Test-Selection Workflow
When the signs alternate:
- Run the term test first.
- Inspect the magnitude sequence.
- Use the Alternating Series Test when magnitudes decrease to zero.
- If the absolute-value series is easy to analyze, keep that information for the absolute/conditional convergence classification.
Common Error: Checking Only the Sign Pattern
Alternating signs alone do not imply convergence.
The series
\[ 1-1+1-1+\cdots \]alternates but diverges because its terms do not approach zero.
Common Error: Forgetting the Limit Condition
If
\[ b_n\not\to0, \]the series diverges regardless of monotonicity.
Common Error: Demanding Decrease from the First Term
The Alternating Series Test requires eventual decrease, not necessarily decrease from the very first index.
Common Error: Treating a Failed Hypothesis as Divergence
If the magnitudes are not decreasing, the Alternating Series Test is inconclusive.
Another method may still prove convergence.
Common Error: Using \(b_N\) Instead of \(b_{N+1}\)
After summing through term \(N\), the first omitted term is term \(N+1\).
Therefore,
\[ |R_N|\le b_{N+1}. \]Common Error: Confusing Term Size with Partial-Sum Error
The error bound works because the series satisfies the alternating-series hypotheses.
It is not a universal rule for arbitrary convergent series.
Common Error: Ignoring the Index Shift
If a series starts at \(n=0\), \(n=2\), or another index, carefully identify which term is actually the first omitted term before applying the remainder bound.
Worked Example: Establish Convergence
Determine whether
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n+4} \]converges.
Let
\[ b_n=\frac1{n+4}. \]Then \(b_n\) decreases and \(b_n\to0\). Therefore the series converges by the Alternating Series Test.
Worked Example: Term Test First
Determine whether
\[ \sum_{n=1}^{\infty}(-1)^n\frac{3n+1}{2n+5} \]converges.
The magnitude satisfies
\[ \frac{3n+1}{2n+5}\to\frac32. \]Therefore the terms do not approach zero, so the series diverges.
Worked Example: Error Requirement
Approximate
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} \]with error less than \(10^{-4}\).
Require
\[ \frac1{(N+1)^2}<10^{-4}. \]Thus
\[ N+1>100. \]So \(N=100\) terms are sufficient.
Worked Example: Eventual Decrease
Consider
\[ \sum_{n=1}^{\infty}(-1)^n\frac{n+10}{n^2+1}. \]The magnitudes approach zero because the denominator has higher degree. For sufficiently large \(n\), the terms decrease. Therefore the alternating series converges.
Worked Example: Bounding the Sum
For a convergent alternating series, suppose
\[ S_8=0.692 \]and
\[ S_9=0.705. \]Then
\[ 0.692\le S\le0.705. \]Practice 1: Basic Alternating Series
Determine whether
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n+1} \]converges.
Practice 2: Term Test
Determine whether
\[ \sum_{n=1}^{\infty}(-1)^n\frac{n}{n+2} \]converges.
Practice 3: Alternating \(p\)-Type Series
Determine whether
\[ \sum_{n=1}^{\infty}\frac{(-1)^n}{n^{2/3}} \]converges.
Practice 4: Logarithmic Magnitudes
Determine whether
\[ \sum_{n=3}^{\infty}\frac{(-1)^{n-1}}{\ln n} \]converges.
Practice 5: Error Bound
For
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}, \]find an upper bound for the error after \(20\) terms.
Practice 6: Number of Terms
How many terms of
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3} \]are sufficient to guarantee error less than \(10^{-5}\)?
Practice 7: Rational Magnitudes
Determine whether
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+4} \]converges.
Practice 8: Identify the Failure
Explain why the Alternating Series Test does not prove convergence of
\[ \sum_{n=1}^{\infty}(-1)^n\left(1+\frac1n\right). \]Practice 9: Bounding the Sum
Suppose a convergent alternating series has
\[ S_{12}=1.428 \]and
\[ S_{13}=1.436. \]Give an interval containing the exact sum.
Practice 10: First Omitted Term
A convergent alternating series has magnitude terms
\[ b_n=\frac1{2n+1}. \]If the partial sum ends at \(n=30\), what alternating-series error bound applies?
Practice Answers
Practice 1
Here
\[ b_n=\frac1{n+1}. \]The sequence decreases and approaches zero, so the series converges by the Alternating Series Test.
Practice 2
Since
\[ \frac{n}{n+2}\to1, \]the terms do not approach zero. The series diverges.
Practice 3
The magnitude
\[ \frac1{n^{2/3}} \]decreases to zero. Therefore the series converges by the Alternating Series Test.
Practice 4
For \(n\ge3\),
\[ b_n=\frac1{\ln n} \]is positive, decreasing, and approaches zero. Therefore the series converges.
Practice 5
The first omitted term after \(20\) terms is
\[ b_{21}=\frac1{21^2}. \]Thus
\[ \boxed{|R_{20}|\le\frac1{441}}. \]Practice 6
Require
\[ \frac1{(N+1)^3}<10^{-5}. \]Since
\[ 10^{5/3}\approx46.42, \]we need \(N+1>46.42\). Thus
\[ \boxed{N=46} \]terms are sufficient.
Practice 7
The magnitude
\[ b_n=\frac{n}{n^2+4} \]approaches zero and is eventually decreasing. Therefore the series converges by the Alternating Series Test.
Practice 8
The magnitudes satisfy
\[ 1+\frac1n\to1, \]not zero. Therefore the series diverges by the term test.
Practice 9
The exact sum lies between consecutive partial sums:
\[ \boxed{1.428\le S\le1.436}. \]Practice 10
The first omitted term is \(n=31\):
\[ b_{31}=\frac1{2(31)+1}=\frac1{63}. \]Therefore
\[ \boxed{|R_{30}|\le\frac1{63}}. \]What to Remember
For a series of the form
\[ \sum (-1)^{n-1}b_n, \]the Alternating Series Test proves convergence when:
- \(b_n\ge0\);
- \(b_n\) is eventually decreasing;
- \(b_n\to0\).
When those conditions hold, the truncation error satisfies
\[ \boxed{|R_N|\le b_{N+1}}. \]The practical habit is to check the term limit first, then monotonicity, and only then use the alternating-series conclusion.
Where We Go Next
Alternating signs can produce convergence through cancellation, but not every convergent alternating series behaves the same way.
The next lesson, Absolute and Conditional Convergence, separates series that remain convergent after removing their signs from those that depend essentially on cancellation.