Alternating Series Test for Convergence
Introduced| English | Meaning |
|---|---|
| alternating series/ˈɔːltəneɪtɪŋ ˈsɪəriːz/ | A series whose successive nonzero terms alternate between positive and negative signs. |
| partial sum/ˈpɑːʃl sʌm/ | partial sum |
| term magnitude | term magnitude |
| decreasing/ˈdiːkriːsɪŋ/ | Describing a function or sequence whose values fall as its input or index increases over a specified range. |
Can repeated adjustments settle?
- A model adds $1$, subtracts $1/2$, adds $1/3$, subtracts $1/4$, and continues. The running total moves up and down instead of only increasing.
- An alternating series 交错级数 has signs that switch. Cancellation can help its partial sums converge, but sign changes alone are not a convergence proof.
Separate the sign from the magnitude
- Write $a_n=(-1)^{n-1}b_n$ with a positive term magnitude 项的绝对值 $b_n$. This sign convention starts positive at $n=1$; using $(-1)^n$ starts negative instead.
- Check the conditions on $b_n$, not on the signed terms $a_n$. For the alternating harmonic series, $b_n=1/n$.
The Alternating Series Test checks decrease and the zero limit on the positive magnitudes, not on the signed terms.
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First isolate b_n from the alternating sign.
For $a_n=(-1)^n/n$, what is $a_1$?
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The first sign is negative since (−1)^1=−1.
State sufficient conditions on a tail
- The Alternating Series Test proves convergence if, from some index onward, $b_n>0$, $b_{n+1}\le b_n$ and $b_n\to0$. Here, decreasing 递减 allows equal consecutive values.
- These are sufficient conditions, not a complete description of every convergent alternating series. Finitely many finite early terms do not change convergence.
For positive magnitudes, which conditions are sufficient for the Alternating Series Test? Select all.
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Both conditions need only hold on a valid tail.
Every convergent alternating series must have eventually non-increasing magnitudes.
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The theorem gives sufficient conditions. A different argument can prove convergence when monotonicity fails.
Changing finitely many finite early terms leaves convergence unchanged.
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The later partial sums change by a fixed finite amount.
Why the partial sums approach one limit
- In the positive-first case with decreasing magnitudes from the start, the even partial sums 部分和 increase and the odd partial sums decrease. Each even total lies below the preceding odd total, so both groups are bounded.
- Their gap is $S_{2k-1}-S_{2k}=b_{2k}\to0$. The two groups therefore approach the same finite limit. Applying this argument to a valid tail gives the eventual version too.
Under the positive-first test conditions, the gap $S_{2k-1}-S_{2k}=b_{2k}$ tends to ____.
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The odd and even partial sums therefore have the same limit.
Apply the test to the alternating harmonic series
- For $\sum_{n=1}^{\infty}(-1)^{n-1}/n$, the magnitudes $b_n=1/n$ are positive, satisfy $1/(n+1)<1/n$, and tend to zero. The series converges by the Alternating Series Test.
- The positive harmonic series $\sum1/n$ diverges. A convergence test for positive series cannot simply be applied to the signed series as if its signs were absent.
Magnitudes and the two groups of partial sums
The default alternating harmonic case has decreasing magnitudes and odd/even partial sums approaching one limit. Other cases distinguish eventual decrease, a nonzero term limit and convergence by a separate pairing argument.
The series $\sum_{n=1}^{\infty}(-1)^{n-1}/n$...
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The positive magnitudes 1/n decrease to zero.
$\sum(-1)^n/n$ converges although $\sum 1/n$ diverges.
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The signed series and its positive-magnitude series require separate arguments.
Find the first four-term partial sum $1-1/2+1/3-1/4$.
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Use denominator 12:12−6+4−3=7, giving 7/12.
Starting with a minus sign changes the total's sign
- The series $\sum_{n=1}^{\infty}(-1)^n/n$ is the negative of the positive-first alternating harmonic series. It also converges; every partial sum is multiplied by $-1$.
- Do not identify either series sum with the individual term limit, which is zero. The test establishes existence of a finite sum, without calculating its exact value.
Multiplying every term of a convergent series by −1 makes it diverge.
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Every partial sum changes sign and still approaches a finite limit.
The term limit zero is the exact infinite sum of every convergent alternating series.
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The limit of individual terms differs from the limit of the partial sums.
Early increases do not spoil a valid tail
- For $b_n=n/(n^2+9)$, use $f(x)=x/(x^2+9)$. It is positive, and $f'(x)=(9-x^2)/(x^2+9)^2\le0$ for $x\ge3$.
- Also $b_n=(1/n)/(1+9/n^2)\to0$. Thus $\sum_{n=1}^{\infty}(-1)^{n-1}n/(n^2+9)$ converges: the conditions hold from $n=3$, even though the first magnitudes increase.
For $f(x)=x/(x^2+9)$, the lesson uses $f\prime(x)\le0$ for all x≥N. Enter the integer N used.
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The derivative numerator 9−x² is non-positive for x≥3.
An increase among the first two magnitudes always prevents use of the Alternating Series Test.
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The conditions only need to hold eventually, after finitely many finite terms.
If the magnitudes do not approach zero
- In $\sum_{n=1}^{\infty}(-1)^{n-1}(1+1/n)$, the positive magnitudes decrease but tend to $1$, not zero.
- The signed terms therefore do not tend to zero. The series diverges by the nth Term Test; the sign alternation does not rescue it.
If positive magnitudes do not tend to zero, the alternating series...
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Its signed terms also fail to tend to zero.
Find $\lim_{n\to\infty}(1+1/n)$.
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The magnitudes decrease to 1, rather than to zero.
Decreasing positive magnitudes alone guarantee alternating-series convergence.
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A positive nonzero magnitude limit fails the necessary zero-term condition.
Failed monotonicity is not a divergence verdict
- Consider $b_{2k-1}=1/k$ and $b_{2k}=1/k+1/k^2$. Every odd-to-even step increases the magnitude, so no tail is non-increasing. The Alternating Series Test has not been justified.
- Yet the positive-first signed series converges by another argument: each adjacent pair adds to $-1/k^2$. Its even partial sums converge, and the difference to the next odd partial sum tends to zero. Pairing here keeps the original order.
If no tail of the magnitudes is non-increasing, the Alternating Series Test proves divergence.
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The theorem has not been justified; failure of this sufficient condition is not a divergence proof.
In the paired example, each signed pair adds to $-1/k^2$. What is the second pair total, at k=2?
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The pair is 1/2−(1/2+1/4)=−1/4.
Why does the non-monotone paired example converge?
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The even partial sums converge by the p-series theorem, and the next odd total differs by a term tending to zero.
Keep the conclusion within the test's scope
- A proved zero limit alone does not establish convergence. A failure of the decrease condition alone does not establish divergence. Choose another argument when the test is not justified.
- The test does not by itself classify the magnitude series or calculate an exact sum. Absolute versus conditional convergence is developed in lesson 10.9; numerical error bounds are in lesson 10.10.
The Alternating Series Test by itself decides whether the series of absolute values converges.
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The magnitude series needs its own convergence argument.
Practise three different condition checks
- Test $\sum_{n=1}^{\infty}(-1)^{n-1}2/\sqrt{n+3}$ and $\sum_{n=1}^{\infty}(-1)^n n/(n^2+16)$. For the second, identify a tail where the magnitudes are non-increasing.
- Then study $\sum_{n=1}^{\infty}(-1)^{n-1}(3n+1)/(n+2)$. State the magnitude limit and name the test that settles the result.
Test $\sum_{n=1}^{\infty}(-1)^{n-1}2/\sqrt{n+3}$. Justify both magnitude conditions.
Compare the magnitude conditions and the named test with the model.
Test $\sum_{n=1}^{\infty}(-1)^n n/(n^2+16)$. Give a valid tail for the magnitude check.
Compare the magnitude conditions and the named test with the model.
Test $\sum_{n=1}^{\infty}(-1)^{n-1}(3n+1)/(n+2)$. State which test supplies the conclusion.
Compare the magnitude conditions and the named test with the model.
Write the conditions before the conclusion
- Identify the positive magnitudes, justify eventual non-increase and show their limit is zero. Then conclude convergence by the Alternating Series Test.
- If a condition fails, explain exactly what is known. Use the nth Term Test for terms that do not approach zero, and another argument when only monotonicity is missing.
Eventually positive, non-increasing magnitudes tending to zero are sufficient for alternating-series convergence. Check the magnitudes rather than the signs. A finite plot illustrates the partial sums; it does not replace the condition checks or calculate the infinite sum.
Which statements correctly describe a complete argument? Select all.
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The sufficient conditions concern a valid tail of the positive magnitudes.