True or False? If the terms of a series tend to zero, the series converges.

False. The test works in one direction only. The harmonic series has terms tending to zero and still diverges, so another test is needed whenever the limit is zero.

True or False? A series smaller than a divergent series must also diverge.

False. The inequality has to point the right way. Being below a divergent series says nothing, just as being above a convergent one says nothing.

What must be true of the terms before either comparison test can be used?

They must be non-negative. Both tests compare sizes term by term, which only makes sense when nothing is negative.

True or False? If an alternating series fails the alternating series test, it diverges.

False. The test can only confirm convergence. There are convergent alternating series whose terms are not non-increasing, so failing the test settles nothing.

An alternating series passes the test. What can you then find?

An error bound. Once convergence is confirmed, the size of the next term bounds how far any partial sum is from the true value.

Define what it means for a series to converge conditionally .

The series converges, but its series of absolute values does not. The alternating harmonic series is the standard example.

True or False? A series that converges absolutely also converges.

True. Absolute convergence is the stronger property, so it settles ordinary convergence at once and the series needs no separate test.

The series of absolute values diverges. What does that tell you?

Nothing on its own. The original series may still converge conditionally, or it may diverge, so it has to be tested separately.

Why does it matter whether a convergent series converges absolutely or conditionally?

Rearranging or regrouping an absolutely convergent series leaves its value unchanged. A conditionally convergent series can be rearranged to give a different sum, or none at all.

What must you check before using the alternating series error bound?

That the series actually converges. Use the alternating series test, or show that it converges absolutely, before bounding anything.

How do the partial sums of a convergent alternating series behave?

They fall alternately above and below the true sum, closing in on it each time. That is why the size of the next term bounds the error.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 10: Infinite Sequences and SeriesTests for Divergence & Convergence

Tests for Divergence & Convergence (College Board AP® Calculus BC): Flashcards

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Frontnth Term Test for Divergence
The $n$ th term test shows that a series diverges. For $\sum_{n = 1}^{\infty} a_{n}$ , the series diverges if
$\lim_{n \to \infty} a_{n}______0$

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Cards in this collection (43)

The $n$ th term test shows that a series diverges. For $\sum_{n = 1}^{\infty} a_{n}$ , the series diverges if
$\lim_{n \to \infty} a_{n}______0$
The completed condition is $\lim_{n \to \infty} a_{n} \neq 0$ .
If the terms do not shrink towards zero then the partial sums cannot settle, so the series must diverge.
True or False?
If the terms of a series tend to zero, the series converges.
False.
The test works in one direction only.
The harmonic series has terms tending to zero and still diverges, so another test is needed whenever the limit is zero.
Why is the $n$ th term test a good first test to apply?
It needs only the limit of a single term, so it is quick.
If that limit is not zero the question is settled at once and no other test is needed.
Determine whether $\sum_{n = 1}^{\infty} \frac{3 n^{2}}{1 - 5 n^{2}}$ converges or diverges.
Dividing top and bottom by $n^{2}$ gives $\frac{3}{\frac{1}{n^{2}} - 5}$ , whose limit is $- \frac{3}{5}$ .
That is not zero, so the series diverges.
Which single condition must every convergent series satisfy?
Its terms must tend to zero.
That is the $n$ th term test read the other way round, and it is why a limit of zero is necessary but never sufficient.
The integral test needs $a_{n} = f (n)$ where $f$ is continuous, positive and $______$ on the interval, so that the rectangles sit neatly against the curve.
The completed condition is that $f$ must be continuous, positive and decreasing on the interval.
A function that rises somewhere on the interval breaks the comparison between the rectangles and the area beneath the curve.
What does the integral test conclude?
The series $\sum_{n = c}^{\infty} a_{n}$ converges exactly when the improper integral $\int_{c}^{\infty} f (x) d x$ exists.
If that integral diverges, so does the series.
Why does a finite area under the curve force the series to converge?
Each term is the area of a rectangle of width $1$ and height $a_{n}$ , and those rectangles are trapped between $A$ and $A + a_{1}$ .
A finite $A$ therefore bounds the whole sum.
True or False?
If the integral test applies, the series has the same value as the integral.
False.
The test decides only whether the series converges, not what it sums to.
The sum lies between $A$ and $A + a_{1}$ , so the integral bounds it rather than equalling it.
Given that $\int_{1}^{\infty} \frac{1}{x} d x$ diverges, what does the integral test say about $\sum_{n = 1}^{\infty} \frac{1}{n}$ ?
That the harmonic series diverges too.
The function $\frac{1}{x}$ is continuous, positive and decreasing for $x \geq 1$ , so the test applies and the two behave alike.
Given that $\int_{1}^{\infty} \frac{1}{x^{2}} d x = 1$ , what does the integral test say about $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ ?
That it converges, since the integral exists with a finite value.
Note that the series does not sum to $1$ : the integral only bounds it.
How does the comparison test show a series converges?
Find a convergent series $\sum b_{n}$ with $a_{n} \leq b_{n}$ for all $n$ .
Being trapped below a convergent series forces $\sum a_{n}$ to converge as well.
How does the comparison test show a series diverges?
Find a divergent series $\sum b_{n}$ with $a_{n} \geq b_{n}$ for all $n$ .
Sitting above a divergent series forces $\sum a_{n}$ to diverge as well.
True or False?
A series smaller than a divergent series must also diverge.
False.
The inequality has to point the right way.
Being below a divergent series says nothing, just as being above a convergent one says nothing.
What must be true of the terms before either comparison test can be used?
They must be non-negative.
Both tests compare sizes term by term, which only makes sense when nothing is negative.
In the limit comparison test the two series behave alike when the limit of their quotient is positive and finite:
$\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L$ where $0 < L <______$
The completed condition is $\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L$ where $0 < L < \infty$ .
When the limit is positive and finite the two series either both converge or both diverge, so the known one settles the unknown one.
A series has terms that are a rational function of $n$ . How do you choose a comparison series?
Keep only the highest power in the numerator and in the denominator, ignoring coefficients.
For $\frac{3 n^{2} + 2 n + 7}{5 n^{4} + n^{3} + 14}$ that gives $\frac{1}{n^{2}}$ , a convergent $p$ -series.
Use a comparison to decide $\sum_{n = 1}^{\infty} \frac{1}{n^{3} + 4}$ .
Since $n^{3} + 4 > n^{3}$ , the terms satisfy $0 < \frac{1}{n^{3} + 4} < \frac{1}{n^{3}}$ .
That is a convergent $p$ -series with $p = 3$ , so the series converges.
Use the limit comparison test on $\sum_{n = 1}^{\infty} \frac{n + 3}{n^{2} + 1}$ .
The highest powers suggest comparing with $\sum \frac{1}{n}$ .
The quotient is $\frac{n^{2} + 3 n}{n^{2} + 1}$ , whose limit is $1$ , so both behave alike and the series diverges.
The ratio test looks at the size of the ratio of consecutive terms:
$\lim_{n \to \infty} |\frac{a_{n + 1}}{______}|$
The completed limit is $\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}|$ .
The next term is divided by the current one, and the limit of that ratio decides the series.
What are the three possible outcomes of the ratio test?
A limit below $1$ means the series converges, and in fact converges absolutely.
A limit above $1$ , or an infinite limit, means it diverges.
A limit of exactly $1$ tells you nothing.
True or False?
A ratio-test limit of exactly $1$ means the series diverges.
False.
The test is simply inconclusive there and another must be used.
Both the convergent $\sum \frac{1}{n^{2}}$ and the divergent $\sum \frac{1}{n}$ give a limit of $1$ .
Which kinds of series is the ratio test best suited to?
Those whose terms contain factorials, exponentials or logarithms, since those simplify sharply when consecutive terms are divided.
It is usually inconclusive for rational functions of $n$ .
Apply the ratio test to $\sum_{n = 1}^{\infty} \frac{n!}{e^{n}}$ .
The ratio simplifies to $\frac{n + 1}{e}$ , since $(n + 1)!$ divided by $n!$ is $n + 1$ .
Its limit is infinite, so the series diverges.
Apply the ratio test to $\sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{{(\ln 2)}^{n}}{n^{4}}$ .
The ratio has magnitude $\ln 2 {(\frac{n}{n + 1})}^{4}$ , whose limit is $\ln 2 = 0.693$ .
That is below $1$ , so the series converges.
Define an alternating series.
A series whose terms alternate in sign, written $\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} a_{n}$ or $\sum_{n = 1}^{\infty} {(- 1)}^{n} a_{n}$ with every $a_{n} > 0$ .
The first form starts positive and the second starts negative.
An alternating series with $a_{n} > 0$ converges when the sizes of the terms never increase and, in addition,
$\lim_{n \to \infty} a_{n} =______$
The completed condition is $\lim_{n \to \infty} a_{n} = 0$ .
Both conditions are needed: the sizes must be non-increasing and they must shrink away to nothing.
True or False?
If an alternating series fails the alternating series test, it diverges.
False.
The test can only confirm convergence.
There are convergent alternating series whose terms are not non-increasing, so failing the test settles nothing.
Show that the alternating harmonic series converges.
Written as $\sum {(- 1)}^{n + 1} \frac{1}{n}$ it has $a_{n} = \frac{1}{n}$ .
The sizes $1, \frac{1}{2}, \frac{1}{3}, \dots$ decrease and their limit is zero, so both conditions hold.
An alternating series passes the test. What can you then find?
An error bound.
Once convergence is confirmed, the size of the next term bounds how far any partial sum is from the true value.
Define what it means for a series to converge absolutely.
The series of absolute values $\sum_{n = 1}^{\infty} |a_{n}|$ converges.
That is a stronger requirement than the series itself converging.
Define what it means for a series to converge conditionally.
The series converges, but its series of absolute values does not.
The alternating harmonic series is the standard example.
True or False?
A series that converges absolutely also converges.
True.
Absolute convergence is the stronger property, so it settles ordinary convergence at once and the series needs no separate test.
The series of absolute values diverges. What does that tell you?
Nothing on its own.
The original series may still converge conditionally, or it may diverge, so it has to be tested separately.
Why does it matter whether a convergent series converges absolutely or conditionally?
Rearranging or regrouping an absolutely convergent series leaves its value unchanged.
A conditionally convergent series can be rearranged to give a different sum, or none at all.
The series $1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \dots$ sums to $0$ . What happens if it is rearranged?
It can be rearranged and regrouped into the alternating harmonic series, which sums to $\ln 2$ .
The series is only conditionally convergent, so its value depends on the order of its terms.
Classify $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n^{2}}$ .
Its absolute values give $\sum \frac{1}{n^{2}}$ , a convergent $p$ -series with $p = 2$ .
So the series converges absolutely, and no separate test is needed.
For a convergent alternating series with sum $S$ and $n$ th partial sum $s_{n}$ , the error is bounded by the next term:
$| S - s_{n} | \leq______$
The completed bound is $|S - s_{n}| \leq a_{n + 1}$ .
The distance from any partial sum to the true value is at most the size of the first term left out.
What must you check before using the alternating series error bound?
That the series actually converges.
Use the alternating series test, or show that it converges absolutely, before bounding anything.
How do the partial sums of a convergent alternating series behave?
They fall alternately above and below the true sum, closing in on it each time.
That is why the size of the next term bounds the error.
True or False?
A partial sum of a convergent alternating series is always an underestimate.
False.
It alternates.
The sign of $S - s_{n}$ matches the sign of the next term, so a partial sum falls short when that term is positive and overshoots when it is negative.
For $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n^{2}}$ , find the least $n$ with $|S - s_{n}| \leq 0.001$ .
Solve $\frac{1}{{(n + 1)}^{2}} < 0.001$ , giving ${(n + 1)}^{2} > 1000$ and $n > 30.62$ .
Since $n$ is a whole number, the least value is $31$ .
The bound you obtain is smaller than the one the question asks for. Is that a problem?
No, it is stronger than needed.
Chain the two together: if $a_{n + 1} = \frac{1}{15}$ and an error below $\frac{1}{10}$ is wanted, write $|S - s_{n}| < \frac{1}{15} < \frac{1}{10}$ .

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