Recognize a p-series

The same terms in two forms

Each dot shows one term of a series, and its height is the value of the term. The term at position \(n\) is \(\frac{1}{n\sqrt{n}}\), and the orange dot shows the term for the \(n\) on the slider.

Press the button. A square root is a power with the exponent one half, so \(\sqrt{n}=n^{1/2}\). When you multiply powers of \(n\), you add the exponents, so \(n\cdot n^{1/2}=n^{3/2}\) and the term becomes \(\frac{1}{n^{3/2}}\).

At \(n=4\), the first form gives \(\frac{1}{4\cdot 2}=\frac18\). The last form gives \(\frac18\) too, because \(4^{3/2}=8\). The dots do not move, because rewriting changes the form of a term and not its value.

The last form, 1 over \(n\) raised to one number, is the form to look for, and here that number is the orange exponent \(\frac32\).

One exponent fits every term

The scene is the same without numbers. The dots now show the terms \(\frac{1}{n^{\cy{p}}}\) for one fixed number \(p\).

Move the slider. A larger \(p\) makes the dots fall faster, because a larger power of \(n\) in the denominator makes the term smaller.

One number \(p\) fits every dot, and only \(n\) changes from one term to the next.

Notation

1. A p-series is a series whose terms are 1 over \(n\) to the power \(p\), where \(p\) is one fixed number.

   \[\sum_{n=1}^{\infty}\frac{1}{n^{\cy{p}}}\]

2. To find \(p\), write each root as a power of \(n\). A root is a power with a fraction as the exponent.

   \[\sqrt[m]{n^{k}}=n^{k/m}\]

3. When you multiply powers of \(n\), you add the exponents. When you divide them, you subtract the exponents.

   \[n^{a}\cdot n^{b}=n^{a+b},\qquad \frac{n^{a}}{n^{b}}=n^{a-b}\]

4. Combine the powers until the term is 1 over one power of \(n\). The exponent in the denominator is \(p\).

   A constant factor in the numerator belongs to the last page of this topic.

When it fails

Take the terms \(\frac{1}{n^{2}+1}\). The faint dots are \(\frac{1}{n^{2}}\), and the orange dots are the new terms. At \(n=1\), the new term is \(\frac12\), and the faint dot is at 1.

Every p-series term equals \(\frac{1}{1^{p}}=1\) at \(n=1\), so no \(p\) fits these terms. A sum in the denominator is not one power of \(n\). The comparison tests in the next topic handle such terms.

Example 1: A root in the denominator

Write \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt[4]{n^{3}}}\) as a p-series and name \(p\).

1. The fourth root of \(n^{3}\) is a power of \(n\) with the exponent \(\frac34\).

   \[\sqrt[4]{n^{3}}=n^{3/4}\]

2. Put this power in the denominator. The term is now 1 over one power of \(n\).

   \[\frac{1}{\sqrt[4]{n^{3}}}=\frac{1}{n^{\cy{3/4}}}\]

3. The series is a p-series with \(p=\frac34\). As a check, at \(n=16\) the power \(16^{3/4}\) is 8, so the term is \(\frac18\), the height of the dot.

   \[16^{3/4}=8\]

Example 2: A power in the numerator

Write \(\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{n}}{n^{2}}\) as a p-series and name \(p\).

1. Write the cube root as a power of \(n\).

   \[\frac{\sqrt[3]{n}}{n^{2}}=\frac{n^{1/3}}{n^{2}}\]

2. A tempting wrong move: The denominator is \(n^{2}\), so \(p=2\).

   \[p=2\]

   The red dots show \(\frac{1}{n^{2}}\), the wrong guess, and they lie below the orange dots of the series. At \(n=8\), the term is \(\frac{2}{64}=\frac1{32}\), and \(\frac{1}{8^{2}}\) is only \(\frac1{64}\). The factor \(\sqrt[3]{n}\) in the numerator makes every term larger.

3. Subtract the exponents, because the power in the numerator cancels part of the power in the denominator.

   \[\frac{n^{1/3}}{n^{2}}=n^{1/3-2}=n^{-5/3}=\frac{1}{n^{\cy{5/3}}}\]

4. The series is a p-series with \(p=\frac53\). At \(n=8\), the power \(8^{5/3}\) is 32, which matches the orange dot.

Practice

Write \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}\sqrt[5]{n}}\) as a p-series and name \(p\).

Write the fifth root as a power of \(n\). Then add the exponents in the denominator. Move the slider until the faint dots lie on the orange dots.

Questions

The dots show the terms of a p-series. The dashed lines mark the height of the dot at \(n=8\) and the height of the dot at \(n=27\).

1. The dot at \(n=8\) has height \(\frac14\). What is \(p\)?

   • \(\frac23\)
   • \(\frac32\)
   • 4
   Answer

   \(\frac23\)

2. Which formula gives the height of the dot at every n?

   • \(\dfrac{1}{\sqrt[3]{n^{2}}}\)
   • \(\dfrac{2}{n}\)
   • \(\dfrac{1}{n^{3/2}}\)
   Answer

   \(\dfrac{1}{\sqrt[3]{n^{2}}}\)

3. Which series is a p-series with \(p=\frac52\)?

   • \(\displaystyle\sum\frac{n}{n^{7/2}}\)
   • \(\displaystyle\sum\frac{1}{n^{2}+n^{1/2}}\)
   • \(\displaystyle\sum\frac{n^{2}}{n^{7/2}}\)
   Answer

   \(\displaystyle\sum\frac{n}{n^{7/2}}\)

Summary

You rewrite each term as 1 over n raised to one fixed power.

The exponent rules change the form of each term and not its value, so the new form describes the same series.

You use this form to match a series with the p-series before you apply the threshold on the next page.

This is one of four skills in topic 10.5 of Which & Why: AP Calculus.