Choose a useful comparison series
For large n, only the leading powers matter
Each dot in the upper graph is one term of a series. The orange dots are the terms \(a_n=\frac{2n+5}{n^{3}+4n}\), and the chalk dots are the simpler terms \(\frac{2}{n^{2}}\). Each line of the scale is ten times as high as the line below it.
The lower graph shows the ratio of the orange term to the chalk term. It falls from 1.20 at n = 10 to 1.002 at n = 1000.
For large \(n\), the number 5 is small next to \(2n\), and \(4n\) is small next to \(n^{3}\). The top behaves like \(2n\) and the bottom behaves like \(n^{3}\), so the term behaves like \(\frac{2n}{n^{3}}=\frac{2}{n^{2}}\).
The series of \(\frac{2}{n^{2}}\) is 2 times a p-series, so you know whether it converges, and this makes it the series to compare with.
The leading term of the top over the leading term of the bottom
The orange dots are the terms \(a_n\), written as a fraction. The top and the bottom each have a leading term, which has the highest power of \(n\). The chalk dots are the leading term of the top divided by the leading term of the bottom.
Move the slider to larger n. The orange dots come closer to the chalk dots and the ratio approaches 1, because the other terms grow more slowly. If you keep only the leading terms, the orange dots move onto the chalk dots.
Notation
Choose a comparison series \(b_n\) so that, for large \(n\), the ratio \(\cy{a_n}/b_n\) approaches a positive number.
In a fraction, keep the leading terms of the top and the bottom.
\[\cy{\frac{2n+5}{n^{3}+4n}}\approx\frac{2n}{n^{3}}=\frac{2}{n^{2}}\]Write a root as a power of \(n\) before you compare the powers.
\[\sqrt{n^{3}}=n^{3/2}\]Among exponential terms, keep the one with the largest base, because it grows fastest.
\[\frac{3^{n}+2n}{5^{n}-1}\approx\frac{3^{n}}{5^{n}}=\left(\tfrac35\right)^{n}\]You must know whether the comparison series converges. A p-series converges if \(p>1\). A geometric series with ratio \(r\) converges if \(|r|<1\).
A constant factor does not change whether a series converges, so use \(\frac{1}{n^{2}}\) in place of \(\frac{2}{n^{2}}\).
When it fails
Take \(a_n=\frac{n+40}{n^{3}}\). The leading term of the top is \(n\), so the comparison series is \(\frac1{n^{2}}\). At \(n=2\) the ratio of the two terms is 21, because 40 is much larger than \(n\) there.
The ratio falls to 2 at n = 40 and to 1.01 at n = 4000, so \(\frac1{n^{2}}\) is still correct. Choose the comparison series from the terms for large n, not from the first few terms.
Example 1: A rational term
Choose a comparison series for \(\displaystyle\sum_{n=1}^{\infty}\frac{4n^{2}+1}{5n^{4}-2}\).
Keep the leading terms of the top and the bottom.
\[\frac{4n^{2}+1}{5n^{4}-2}\approx\frac{4n^{2}}{5n^{4}}=\frac45\cdot\frac{1}{n^{2}}\]The factor \(\frac45\) is a constant, so it does not change whether the series converges. Choose \(b_n=\frac{1}{n^{2}}\), the chalk dots. In the lower graph, the ratio of the terms approaches the dashed line at \(\frac45\).
This is a p-series with p = 2, which converges, so the given series is likely to converge. A comparison test then confirms it.
Example 2: A root in the bottom
Choose a comparison series for \(\displaystyle\sum_{n=1}^{\infty}\frac{n+1}{\sqrt{n^{3}+n}}\).
Write the root as a power of \(n\). For large \(n\), the bottom behaves like \(\sqrt{n^{3}}=n^{3/2}\).
A tempting wrong move: The bottom is about \(n^{3/2}\), so choose \(b_n=\frac{1}{n^{3/2}}\).
\[b_n=\frac{1}{n^{3/2}}\]The red dots show this choice and lie far below the orange dots. At \(n=100\) the term is about 0.10, but \(\frac{1}{100^{3/2}}\) is 0.001. The choice leaves out the top, which grows like \(n\).
Keep the leading terms of both the top and the bottom, and subtract the exponents.
\[\frac{n+1}{\sqrt{n^{3}+n}}\approx\frac{n}{n^{3/2}}=\frac{1}{n^{1/2}}\]For large \(n\) the chalk dots lie on the orange dots, and at \(n=100\) the ratio of the two terms is about 1.01. This is a p-series with \(p=\frac12\), which diverges, so the given series is likely to diverge.
Practice
Choose a comparison series for \(\displaystyle\sum_{n=1}^{\infty}\frac{3^{n}+2n}{5^{n}-1}\), and say whether the given series is likely to converge.
For large \(n\), compare \(3^{n}\) with \(2n\), and \(5^{n}\) with 1. Move the slider until the chalk dots lie on the orange dots.
Solution
For large \(n\), \(3^{n}\) is much larger than \(2n\), and \(5^{n}\) is much larger than 1. Keep the leading terms.
\[\frac{3^{n}+2n}{5^{n}-1}\approx\frac{3^{n}}{5^{n}}=\left(\tfrac35\right)^{n}\]For large \(n\), the chalk dots at \(r=0.6\) lie on the orange dots. This is a geometric series with ratio \(\frac35<1\), so it converges and the given series is likely to converge.
Questions
Questions 1 and 2 use the figure. The orange dots are the terms \(a_n=\frac{3n^{2}+7}{n^{4}+2n}\). The chalk dots are three possible comparison series: A is \(\frac1n\), B is \(\frac{3}{n^{2}}\), and C is \(\frac{1}{n^{4}}\). The lower graph shows \(a_n\) divided by \(\frac{3}{n^{2}}\).
Which row of chalk dots comes together with the orange dots as n grows?
- B, \(\frac{3}{n^{2}}\)
- A, \(\frac1n\)
- C, \(\frac{1}{n^{4}}\)
Answer
B, \(\frac{3}{n^{2}}\)
Where do the dots of the lower graph go as n grows?
- Toward 1
- Toward 3
- Toward 0
Answer
Toward 1
Which series is the most useful comparison series for \(\displaystyle\sum_{n=2}^{\infty}\frac{n^{2}+1}{n^{7/2}-n}\)?
- \(\displaystyle\sum\frac{1}{n^{3/2}}\)
- \(\displaystyle\sum\frac{1}{n^{7/2}}\)
- \(\displaystyle\sum\frac{1}{n^{11/2}}\)
- \(\displaystyle\sum\frac{1}{n^{2}}\)
Answer
\(\displaystyle\sum\frac{1}{n^{3/2}}\)
Summary
You keep the leading terms of the top and the bottom and simplify them to a p-series or a geometric series.
For large n the other terms are small, so the ratio of the two terms approaches a positive number.
You do this before a comparison test, because the test needs a series whose convergence you know.
This is one of four skills in topic 10.6 of Which & Why: AP Calculus.