Calculus BC · Unit 10
Check alternating signs, decreasing size, and a zero limit. Then decide whether the absolute series converges.
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The alternating series test
If the signs alternate, the absolute values decrease, and the terms go to 0, then the series converges.
Absolute convergence
If \(\sum |a_n|\) converges, then \(\sum a_n\) converges absolutely.
Conditional convergence
The series converges, but \(\sum |a_n|\) diverges.
Terms must still go to 0
Alternating signs do not save a series whose terms stay away from 0.
Check the three alternating-series conditions, then check the absolute series.
The signs alternate, \(1/n^2\) decreases, and the terms go to 0.
The alternating series test says the series converges.
\(\sum 1/n^2\) is a convergent p-series.
The series converges absolutely.
Answer: absolutely convergent
The signs alternate, \(1/n\) decreases, and the limit is 0, so the series converges.
\(\sum 1/n\) is the harmonic series, which diverges.
The series converges conditionally.
Answer: conditionally convergent
\(\sum 1/n^3\) converges, since \(p=3>1\).
The original series converges absolutely.
Answer: absolutely convergent
\(\lim_{n\to\infty}\dfrac{n}{n+1}=1\neq 0\)
The terms do not go to 0.
The series diverges.
Answer: diverges
The signs alternate, \(1/\sqrt{n}\) decreases, and the limit is 0, so the series converges.
\(\sum 1/\sqrt{n}\) is a p-series with \(p=1/2\), so it diverges.
The series converges conditionally.
Answer: conditionally convergent
A and B converge, but their absolute series are p-series with \(p\le 1\).
D diverges because the terms do not go to 0.
\(\sum 1/n^4\) converges, so C is absolute.
Answer: C. \(\sum (-1)^n/n^4\)
Absolute values that behave like 1/n, not like 1/n².
Let \(b_n=n/(n^2+1)\).
\(\lim b_n=0\)
The derivative of \(x/(x^2+1)\) is \((1-x^2)/(x^2+1)^2\), which is negative for \(x>1\).
So \(b_n\) eventually decreases, and the alternating series test gives convergence.
Limit comparison of \(b_n\) with \(1/n\) has limit 1, so \(\sum b_n\) diverges.
The series converges conditionally.
Answer: conditionally convergent
\(1/\ln(n+1)\) decreases to 0 because \(\ln(n+1)\) increases to infinity.
The alternating series test says the series converges.
For \(n\ge 1\), \(\ln(n+1)
\(\sum 1/(n+1)\) diverges, so the absolute series diverges.
The series converges conditionally.
Answer: conditionally convergent
\(\left|\dfrac{2^n}{n}\right|\) grows without bound.
The terms do not go to 0.
The series diverges.
Answer: diverges
\(\dfrac{n+1}{n^3}=\dfrac{1}{n^2}+\dfrac{1}{n^3}\)
The absolute series is the sum of two convergent p-series.
The series converges absolutely.
Answer: absolutely convergent
The signs alternate.
\(1/n\) is decreasing.
\(1/n\to 0\).
All three hold, so the series converges.
Answer: alternating, decreasing, limit 0, so it converges
For a positive term, \(|a_n|=a_n\).
\(\sum |a_n|\) is the same series, which converges.
The convergence is absolute.
Answer: absolute
Eventually decreasing is enough. A limit that is not zero is not.
\(\lim_{n\to\infty}\dfrac{n^2}{n^2+1}=1\neq 0\)
The terms do not go to 0.
The series diverges.
Answer: diverges
Limit comparison of the absolute value with \(1/\sqrt{n}\):
\(\lim_{n\to\infty}\dfrac{\sqrt{n}/(n+3)}{1/\sqrt{n}}=\lim_{n\to\infty}\dfrac{n}{n+3}=1\)
The absolute series diverges.
The derivative of \(\sqrt{x}/(x+3)\) has numerator \(3-x\), so the terms decrease for \(n\ge 4\).
The terms also go to 0, so the alternating series test applies to the tail.
The series converges conditionally.
Answer: conditionally convergent
\(\dfrac{2n+1}{n^2}=\dfrac{2}{n}+\dfrac{1}{n^2}\)
The absolute series diverges because \(\sum 2/n\) diverges.
The absolute values decrease to 0 for \(n\ge 1\).
The alternating series test says the series converges.
The convergence is conditional.
Answer: conditionally convergent
\(1/(2n+1)\) decreases to 0, so the alternating series test gives convergence.
Limit comparison of \(1/(2n+1)\) with \(1/n\) has limit \(1/2\).
The absolute series diverges.
The series converges conditionally.
Answer: conditionally convergent
The terms do not go to 0, so the series diverges.
Parentheses change the order of a series that does not converge.
The grouped sum is a different series.
Answer: the series diverges, so regrouping is not valid
The statement is false.
\(\sum (-1)^{n+1}/n\) converges, while \(\sum 1/n\) diverges.
That is conditional convergence.
Answer: false; the alternating harmonic series is a counterexample
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