Calculus BC ยท Unit 10
Look at the limit of the absolute ratio of consecutive terms. A limit of 1 means you need a different test.
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The limit
\(L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\)
Three outcomes
\(L<1\) means absolute convergence. \(L>1\), including an infinite limit, means divergence. \(L=1\) means the test is inconclusive.
Factorials and exponentials
The ratio test is built for \(n!\) and \(c^n\). Most of the fraction cancels.
Do not stop at L = 1
A p-series can give \(L=1\) and still converge. Use another test.
Cancel the ratio until a simple limit remains.
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{2^{n+1}/(n+1)!}{2^n/n!}\)
\(=\dfrac{2}{n+1}\)
The limit is 0.
\(0<1\)
The series converges absolutely.
Answer: converges absolutely
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^2}{2^{n+1}}\cdot\dfrac{2^n}{n^2}\)
\(=\dfrac{1}{2}\left(\dfrac{n+1}{n}\right)^2\)
The limit is \(\dfrac{1}{2}\).
\(\dfrac{1}{2}<1\)
The series converges absolutely.
Answer: converges absolutely
\(\left|\dfrac{a_{n+1}}{a_n}\right|=3\cdot\dfrac{n}{n+1}\)
The limit is 3.
\(3>1\)
The series diverges.
Answer: diverges
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n^2}{(n+1)^2}\)
The limit is 1.
The ratio test is inconclusive.
The series does converge, because it is a p-series with \(p=2\), but that fact does not come from the ratio test.
Answer: inconclusive
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{5}{n+1}\)
The limit is 0, which is less than 1.
The series converges absolutely.
Answer: converges absolutely
\(4>1\)
The series diverges.
Answer: C. divergence
A factorial in the numerator, or a ratio that tends to e.
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)!}{5^{n+1}}\cdot\dfrac{5^n}{n!}\)
\(=\dfrac{n+1}{5}\)
The limit is infinity, which is greater than 1.
The series diverges.
Answer: diverges
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^{n+1}}{(n+1)!}\cdot\dfrac{n!}{n^n}\)
\(=\dfrac{(n+1)^n}{n^n}\)
\(=\left(1+\dfrac{1}{n}\right)^n\)
The limit is \(e\), and \(e>1\).
The series diverges.
Answer: diverges
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^2}{(2n+2)(2n+1)}\)
\(=\dfrac{n+1}{2(2n+1)}\)
The limit is \(\dfrac{1}{4}\).
\(\dfrac{1}{4}<1\)
The series converges absolutely.
Answer: converges absolutely
\(\left|\dfrac{a_{n+1}}{a_n}\right|=4\cdot\left(\dfrac{n}{n+1}\right)^3\)
The limit is 4, which is greater than 1.
The series diverges.
Answer: diverges
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n+1}{n}\cdot\dfrac{1}{3}\)
The limit is \(\dfrac{1}{3}<1\).
The series converges absolutely.
Answer: converges absolutely
The ratio test is inconclusive.
The alternating series test applies: the signs alternate, \(1/n\) decreases, and the limit is 0.
The series converges (conditionally).
Answer: inconclusive by the ratio test; converges by the alternating series test
A ratio that tends to 2/e or 1/e.
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{2\,n^n}{(n+1)^n}\)
\(=\dfrac{2}{\left(1+1/n\right)^n}\)
The limit is \(2/e\).
\(2/e<1\)
The series converges absolutely.
Answer: converges absolutely
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n^n}{(n+1)^n}\)
\(=\dfrac{1}{\left(1+1/n\right)^n}\)
The limit is \(1/e<1\).
The series converges absolutely.
Answer: converges absolutely
The absolute ratio is \(5/(n+1)\).
The limit is 0, which is less than 1.
The series converges absolutely.
Answer: converges absolutely
\(0<\dfrac{1}{n^2+1}<\dfrac{1}{n^2}\)
\(\sum 1/n^2\) converges.
By direct comparison, the series converges.
Answer: converges by comparison
\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^3}{n^3}\cdot\dfrac{1}{n+1}\)
\(=\dfrac{(n+1)^2}{n^3}\)
The limit is 0, which is less than 1.
The series converges absolutely.
Answer: converges absolutely
\(L=1\) gives no conclusion.
The series might converge, as \(\sum 1/n^2\) does, or diverge, as \(\sum 1/n\) does.
Another test is required.
Answer: L = 1 is inconclusive
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