Calculus BC · Unit 10
Compare with a series you already know. The inequality has to point the right way, or the limit has to be a positive finite number.
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Direct comparison, convergence
If \(0\le a_n\le b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
Direct comparison, divergence
If \(a_n\ge b_n\ge 0\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
The wrong direction is silence
A larger divergent series, or a smaller convergent series, tells you nothing.
Limit comparison
If \(\lim a_n/b_n=L\) with \(0
Bound the term by a p-series or a geometric series.
For \(n\ge 1\),
\(0<\dfrac{1}{n^2+n}<\dfrac{1}{n^2}\)
\(\sum 1/n^2\) converges.
By direct comparison, the series converges.
Answer: converges
\(0<\dfrac{1}{n^3+4}<\dfrac{1}{n^3}\)
\(\sum 1/n^3\) converges.
The series converges.
Answer: converges
\(0<\dfrac{1}{2^n+n}<\dfrac{1}{2^n}\)
\(\sum (1/2)^n\) converges.
The series converges.
Answer: converges
For \(n\ge 2\), \(n^2-1\ge \dfrac{1}{2}n^2\), because \(n^2\ge 2\).
Taking reciprocals reverses the inequality.
\(\dfrac{1}{n^2-1}\le \dfrac{2}{n^2}\)
\(\sum 2/n^2\) converges, so the series converges.
Answer: converges
A compares with a divergent series of ones, in the wrong direction.
B compares with the convergent p-series \(\sum 1/n^4\).
C is a lower bound by a divergent series, which would suggest divergence and is also false for large \(n\).
D is true and useless.
B is the proof.
Answer: B. \(1/(n^4+n)\le 1/n^4\)
\(0<\dfrac{n}{n^3+5}<\dfrac{n}{n^3}=\dfrac{1}{n^2}\)
\(\sum 1/n^2\) converges.
The series converges.
Answer: converges
Use a limit when a direct inequality is awkward.
\(\lim_{n\to\infty}\dfrac{1/\sqrt{n^2+1}}{1/n}=\lim_{n\to\infty}\dfrac{n}{\sqrt{n^2+1}}\)
\(=\lim_{n\to\infty}\dfrac{1}{\sqrt{1+1/n^2}}\)
The limit is 1, which is positive and finite.
\(\sum 1/n\) diverges.
The series diverges.
Answer: diverges
\(\lim_{n\to\infty}\dfrac{(n+2)/(n^2+3)}{1/n}=\lim_{n\to\infty}\dfrac{n(n+2)}{n^2+3}\)
\(=\lim_{n\to\infty}\dfrac{n^2+2n}{n^2+3}=1\)
\(\sum 1/n\) diverges.
The series diverges.
Answer: diverges
\(0<\dfrac{3^n}{4^n+1}<\left(\dfrac{3}{4}\right)^n\)
\(\sum (3/4)^n\) is a convergent geometric series.
The series converges.
Answer: converges
The harmonic term is smaller than a divergent term.
Divergence of a larger series says nothing about the smaller one.
The harmonic series does diverge, but this comparison does not prove it.
Answer: the inequality runs the wrong way for a divergence proof
\(\lim_{n\to\infty} n^2\cdot\dfrac{2n+1}{n^3+n}=\lim_{n\to\infty}\dfrac{2n^3+n^2}{n^3+n}=2\)
The limit is positive and finite.
\(\sum 1/n^2\) converges.
The series converges.
Answer: converges
\(\lim_{n\to\infty}\dfrac{1/(\sqrt{n}+1)}{1/\sqrt{n}}=\lim_{n\to\infty}\dfrac{\sqrt{n}}{\sqrt{n}+1}=1\)
\(\sum 1/\sqrt{n}\) diverges.
The series diverges.
Answer: diverges
A term that needs an equivalent, or a comparison that looks convincing and is not.
\(\lim_{n\to\infty}\dfrac{\sin(1/n)/n}{1/n^2}=\lim_{n\to\infty} n\sin(1/n)\)
\(=\lim_{n\to\infty}\dfrac{\sin(1/n)}{1/n}=1\)
\(\sum 1/n^2\) converges.
The series converges.
Answer: converges
\(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\)
\(\lim_{n\to\infty}\dfrac{1/(\sqrt{n+1}+\sqrt{n})}{1/\sqrt{n}}=\lim_{n\to\infty}\dfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\)
The limit is \(\dfrac{1}{2}\).
\(\sum 1/\sqrt{n}\) diverges.
The series diverges.
Answer: diverges
Compare with \(1/n^2\).
\(\lim_{n\to\infty}\dfrac{n^2/(n^4+1)}{1/n^2}=\lim_{n\to\infty}\dfrac{n^4}{n^4+1}=1\)
The series converges.
Answer: converges
Being larger than a convergent series gives no conclusion.
The useful bound is \(0<\dfrac{1}{n^2+1}<\dfrac{1}{n^2}\).
\(\sum 1/n^2\) converges, so \(\sum 1/(n^2+1)\) converges.
Answer: the series converges; the student’s direction does not prove divergence
For \(n\ge 1\),
\(0<\dfrac{1}{n\cdot 2^n}\le \dfrac{1}{2^n}\)
\(\sum (1/2)^n\) converges.
The series converges.
Answer: converges
\(\lim_{n\to\infty}\dfrac{n(n^2+1)}{n^3+4n}=\lim_{n\to\infty}\dfrac{n^3+n}{n^3+4n}=1\)
\(\sum 1/n\) diverges.
The series diverges.
Answer: diverges
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