Calculus BC ยท Unit 10
A p-series converges only when p is greater than 1. The integral test needs a positive, continuous, decreasing function.
p-series
\(\sum \dfrac{1}{n^p}\) converges when \(p>1\) and diverges when \(p\le 1\).
The harmonic series
\(\sum \dfrac{1}{n}\) is the case \(p=1\). It diverges, even though the terms go to 0.
Integral test
If \(f\) is positive, continuous, and decreasing, then \(\sum f(n)\) and \(\int_1^{\infty} f(x)\,dx\) either both converge or both diverge.
A constant multiple
Multiplying by a nonzero constant does not change convergence. \(\sum 7/n^2\) converges because \(\sum 1/n^2\) does.
Read the exponent and compare it with 1.
This is a p-series with \(p=3\).
\(3>1\)
The series converges.
Answer: converges
\(\dfrac{1}{\sqrt{n}}=\dfrac{1}{n^{1/2}}\)
\(p=\dfrac{1}{2}\le 1\)
The series diverges.
Answer: diverges
This is the harmonic series, \(p=1\).
The series diverges.
Answer: diverges
\(\dfrac{n}{n^4}=\dfrac{1}{n^3}\)
\(p=3>1\)
The series converges.
Answer: converges
\(\sum \dfrac{1}{n^2}\) converges, since \(p=2>1\).
A constant multiple of a convergent series converges.
Answer: converges
Convergence needs \(p>1\).
\(0.9<1\)
B diverges. The others converge.
Answer: B. \(\sum 1/n^{0.9}\)
Match a series with an improper integral, or rewrite the general term first.
\(\int_1^{\infty} x^{-2}\,dx\)
\(\lim_{b\to\infty}\left[-\dfrac{1}{x}\right]_1^{b}\)
\(\lim_{b\to\infty}\left(-\dfrac{1}{b}+1\right)\)
The limit is 1, so the integral converges.
The series converges.
Answer: converges
\(\int_1^{\infty}\dfrac{1}{x}\,dx\)
\(\lim_{b\to\infty}\left[\ln x\right]_1^{b}\)
\(\lim_{b\to\infty}\ln b\)
The limit is infinite, so the integral diverges.
The series diverges.
Answer: diverges
\(p=\dfrac{4}{3}\)
\(\dfrac{4}{3}>1\)
The series converges.
Answer: converges
\(\dfrac{n^2}{n^5}=\dfrac{1}{n^3}\)
\(p=3>1\)
The series converges.
Answer: converges
\(\int_1^{\infty} e^{-x}\,dx\)
\(\lim_{b\to\infty}\left[-e^{-x}\right]_1^{b}\)
\(\lim_{b\to\infty}\left(-e^{-b}+e^{-1}\right)\)
The limit is \(1/e\), so the integral converges.
The series converges.
Answer: converges
The integral test needs a function that is positive, continuous, and decreasing.
\(\sin x\) changes sign and is not decreasing on \([1,\infty)\).
The hypotheses fail, so the test does not apply.
Answer: sine is not positive and decreasing
An integral with a substitution, or a series that only looks like a p-series.
\(\int_2^{\infty}\dfrac{1}{x\ln x}\,dx\)
Let \(u=\ln x\). Then \(du=\dfrac{1}{x}\,dx\).
The limits run from \(\ln 2\) to \(\infty\).
\(\int_{\ln 2}^{\infty}\dfrac{1}{u}\,du\)
\(\lim_{b\to\infty}\left[\ln u\right]_{\ln 2}^{b}=\infty\)
The integral diverges, so the series diverges.
Answer: diverges
\(\sum 1/n^2\) converges and \(\sum 1/n^5\) converges.
The sum of two convergent series converges.
Answer: converges
\(\dfrac{n+1}{n^3}=\dfrac{1}{n^2}+\dfrac{1}{n^3}\)
Both series on the right are p-series with \(p>1\).
The series converges.
Answer: converges
\(\int_1^{\infty}\dfrac{x}{x^2+1}\,dx\)
Let \(u=x^2+1\). Then \(du=2x\,dx\).
\(\dfrac{1}{2}\int_2^{\infty}\dfrac{1}{u}\,du\)
\(\dfrac{1}{2}\lim_{b\to\infty}\left[\ln u\right]_2^{b}=\infty\)
The integral diverges, so the series diverges.
Answer: diverges
It is not a p-series, because the denominator is not a pure power of \(n\).
\(\int_1^{\infty}\dfrac{1}{x^2+4}\,dx\)
\(\left[\dfrac{1}{2}\arctan\dfrac{x}{2}\right]_1^{\infty}\)
\(\dfrac{1}{2}\left(\dfrac{\pi}{2}-\arctan\dfrac{1}{2}\right)\)
The integral is finite, so the series converges.
Answer: not a p-series, and it converges
\(\dfrac{5}{n^{1/3}}\)
\(p=\dfrac{1}{3}\le 1\)
The series diverges.
Answer: diverges
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