Apply the p-series threshold
Two totals that behave differently
The bars in the upper graph show the terms \(\frac{1}{n^{2}}\) of a series. The teal curve in the lower graph shows the running total, which is the sum of the first \(N\) terms.
Drag \(N\) from 1 to 100. The bars shrink quickly, and the running total rises by less each time. At \(N=100\), it is about 1.63.
Press the button for \(p=\frac12\). The terms are now \(\frac{1}{\sqrt{n}}\). They also shrink, but the running total continues to rise, and at \(N=100\) it is about 18.6.
In both cases the terms shrink, but only one running total levels off. The exponent decides which one.
Rectangles and the area under a curve
The curve is \(y=\frac{1}{x^{p}}\) for \(x\ge1\), and the slider changes \(p\). Each teal rectangle is one unit wide and as tall as the curve at its right end. Its area is therefore one term of the series, from \(n=2\) on.
The curve falls, so each rectangle lies under it, and the terms from \(n=2\) on add to less than the area under the curve.
Turn on the toggle. Each rectangle is now as tall as the curve at its left end, so it lies above the curve. The first \(N\) terms add to more than the area from 1 to \(N+1\).
The sum stays close to the area under the curve. If that area is finite, the sum is finite. If that area has no end, the sum has no end.
Notation
For a positive, decreasing function \(f\), the integral test says that the series and the integral from 1 on are both finite or both infinite.
For \(f(x)=\frac{1}{x^{p}}\) and \(p\ne1\), the area from 1 to \(b\) has this value.
\[\int_{1}^{b}\frac{dx}{x^{p}}=\frac{b^{1-p}-1}{1-p}\]If \(p>1\), the exponent \(1-p\) is negative, so \(b^{1-p}\) approaches 0 as \(b\) grows. The area is finite, and the series converges.
\[\int_{1}^{\infty}\frac{dx}{x^{p}}=\cb{\frac{1}{p-1}}\]If \(0\lt p<1\), the exponent \(1-p\) is positive, so \(b^{1-p}\) grows without bound. If \(p=1\), the area is \(\ln b\), which also grows without bound. In both cases the area has no end, and the series diverges.
\[\int_{1}^{b}\frac{dx}{x}=\ln b\]If \(p\le0\), the terms do not approach 0, so the series diverges.
Together, the cases give the threshold, with \(p=1\) on the divergent side.
\[\sum_{n=1}^{\infty}\frac{1}{n^{p}}\ \begin{cases}\text{converges if }p>1\\ \text{diverges if }p\le 1\end{cases}\]
When it fails
Set \(p=0\). The curve is flat at height 1, so every term equals 1 and the total grows by 1 at each step. The terms do not approach 0, so the series diverges, and the same happens for every \(p<0\).
For \(0
Example 1: An exponent above 1
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{5/4}}\) converges.
The term is already one power of \(n\), and \(p=\frac54\) is greater than 1.
\[p=\frac54>1\]The area under the curve from 1 on is finite.
\[\int_{1}^{\infty}\frac{dx}{x^{5/4}}=\frac{1}{5/4-1}=\cb{4}\]The teal rectangles stand for the terms from \(n=2\) on, and they lie under the curve, so these terms add to less than 4. The first term is 1, so the whole sum is at most 5.
\[\sum_{n=1}^{\infty}\frac{1}{n^{5/4}}\le 1+4=5\]The running total rises and never passes 5, so it approaches a limit. The series converges.
Example 2: An exponent below 1
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[4]{n}}{n}\) converges.
Subtract the exponents to write the term as one power of \(n\).
\[\frac{n^{1/4}}{n}=\frac{1}{n^{\cy{3/4}}}\]A tempting wrong move: The terms \(\frac{1}{n^{3/4}}\) approach 0, so the series converges.
\[\lim_{n\to\infty}\frac{1}{n^{3/4}}=0\ \Rightarrow\ \text{converges}\]The rectangles lie above the curve, so the first \(N\) terms add to more than the area from 1 to \(N+1\). That area is \(4\bigl((N+1)^{1/4}-1\bigr)\), and it has no end. For \(N=1\,000\,000\), it is already more than 120.
The exponent \(p=\frac34\) is not greater than 1, so the area under the curve has no end. The running total is larger than that area, so the series diverges.
Practice
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}\sqrt[6]{n}}\) converges.
Write the sixth root as a power of \(n\). Subtract the exponents to find \(p\), and compare \(p\) with 1. The slider shows the curve for each exponent.
Solution
Write the root as a power of \(n\) and combine the powers.
\[\frac{n^{3}}{n^{4}\cdot n^{1/6}}=n^{3-4-1/6}=\frac{1}{n^{\cy{7/6}}}\]The exponent \(p=\frac76\) is greater than 1, so the area under \(\frac{1}{x^{7/6}}\) is finite, and the series converges.
\[\int_{1}^{\infty}\frac{dx}{x^{7/6}}=\frac{1}{7/6-1}=\cb{6}\]
Questions
The graph shows the running totals of two p-series. One has \(p=\frac32\) and the other has \(p=\frac23\). The orange dots mark the totals at one value of \(N\).
Which curve shows the running total of the series with \(p=\frac32\)?
- Curve A, because its total slows down and has a bound
- Curve B, because a larger exponent makes larger terms
- Curve B, because its total reaches the higher value
Answer
Curve A, because its total slows down and has a bound
Curve B rises by a smaller step each time \(N\) grows. What does this show about its series?
- Nothing about convergence by itself. The series diverges because \(p=\frac23\) is not greater than 1.
- The series converges, because the steps shrink.
- The series converges to about 7.9, the height at N = 40.
Answer
Nothing about convergence by itself. The series diverges because \(p=\frac23\) is not greater than 1.
For which values of \(p\) does \(\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{n}}{n^{p}}\) converge?
- \(p>\frac32\)
- \(p>1\)
- \(p>\frac12\)
Answer
\(p>\frac32\)
Summary
You write the series as a p-series and compare its exponent p with 1.
The sum stays close to the area under \(y=\frac1{x^{p}}\), and that area is finite only when \(p\) is greater than 1.
The threshold decides any p-series at once, and a p-series often serves as the known series in a comparison.
This is one of four skills in topic 10.5 of Which & Why: AP Calculus.