Form the absolute term ratio
Comparing a term with the one before
The upper graph shows the first seven terms of the series \(\sum (-1)^n \frac{5n}{3^n}\), with green bars for positive terms and red bars for negative terms. The lower graph shows the length of each bar, which is its distance from the axis, so it has no sign.
Drag the slider to choose \(n\), and the orange arrow joins bar \(n\) to the next bar. Their ratio, the next length divided by this length, is the factor by which the terms shrink or grow in one step.
This ratio is 2/3 from the first bar to the second, then 1/2, then 4/9. Each bar is shorter than the one before it, but the factor changes each time.
In a geometric series the factor is the same at every step. Here it changes, so the ratio test uses its value for large n.
The ratio of consecutive terms
Drag the slider to choose a term \(a_n\). The ratio \(\cy{\left|\frac{a_{n+1}}{a_n}\right|}\) is the length of the next bar divided by the length of this bar.
The faint dashed line marks the length of this bar. If the ratio is less than 1, the next bar ends below the line. If the ratio is more than 1, the next bar ends above it, as it does when you make the next bar longer.
The absolute value removes the signs, so the colors of the bars do not change the ratio.
Notation
To write the next term, replace every \(n\) in the formula with \(n+1\), including inside factorials and exponents.
\[a_{n+1}\]Divide the next term by the current term. You cannot divide by 0, so every term must be nonzero.
\[\frac{a_{n+1}}{a_n}\]Take the absolute value. A sign factor such as \((-1)^n\) has absolute value 1, so it disappears from the ratio.
\[\cy{\left|\frac{a_{n+1}}{a_n}\right|}\]The ratio test uses the limit \(L\) of this ratio as \(n\) grows. The next skills find that limit.
\[L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|\]
When it fails
In an alternating series, each bar lies on the other side of the axis from the one before it. Without the absolute value, the ratio is negative at every step, as the red arrow in the upper graph shows.
A negative ratio is less than 1 even when the next bar is longer, so it cannot tell you whether the terms grow or shrink. The absolute value compares only the lengths.
Example 1: A power over a linear factor
Form \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} \frac{4^n}{n+1}\). Do not simplify it.
Replace \(n\) with \(n+1\) in every place. The sign factor, the power and the denominator all change.
\[a_{n+1} = (-1)^{n+2} \frac{4^{n+1}}{n+2}\]Divide \(a_{n+1}\) by \(a_n\).
\[\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+2}\, 4^{n+1}/(n+2)}{(-1)^{n+1}\, 4^n/(n+1)}\]Take the absolute value. The sign factors have absolute value 1, so they disappear.
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{4^{n+1}/(n+2)}{4^n/(n+1)}\]Test the formula at \(n = 1\). The first two bars in the lower graph have lengths 2 and \(\frac{16}{3}\), and their ratio matches the formula.
\[\frac{16/3}{2} = \frac{4^{2}/3}{4/2} = \cy{\frac{8}{3}}\]
Example 2: A factorial of an expression in n
Form \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{16^n}{(2n+1)!}\). Do not simplify it.
Replace \(n\) with \(n+1\) in the power and in the factorial of the term \(a_n\).
A tempting wrong move: The factorial \((2n+1)!\) becomes \((2n+2)!\).
\[a_{n+1} = (-1)^{n+1} \frac{16^{n+1}}{(2n+2)!}\]At \(n = 1\), the next term is \(a_2\), whose factorial is \((2 \cdot 2+1)! = 5!\), but \((2n+2)!\) gives \(4!\). The red outline bar shows the wrong length \(\frac{32}{3}\), far above the true length \(\frac{32}{15}\).
Replace \(n\) with \(n+1\) inside the factorial: \(2(n+1)+1 = 2n+3\). The new factorial is \((2n+3)!\).
\[a_{n+1} = (-1)^{n+1} \frac{16^{n+1}}{(2n+3)!}\]Divide and take the absolute value. At \(n = 1\), the formula gives \(\frac45\), the ratio of the first two lengths.
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{16^{n+1}/(2n+3)!}{16^n/(2n+1)!}\]
Practice
Form \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^{2n}}{(n+1)!}\). Test your formula at \(n = 1\).
Replace \(n\) with \(n+1\) in the exponent and in the factorial. At \(n = 1\), your ratio must equal the length of the second bar divided by the length of the first.
Solution
In the next term, the exponent \(2n\) becomes \(2(n+1) = 2n+2\), and the factorial \((n+1)!\) becomes \((n+2)!\).
\[a_{n+1} = (-1)^{n+2} \frac{3^{2n+2}}{(n+2)!}\]Divide, and take the absolute value.
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{3^{2n+2}/(n+2)!}{3^{2n}/(n+1)!}\]Test at \(n = 1\). The first two bars have lengths \(\frac92\) and \(\frac{27}{2}\), so their ratio is 3, and the formula gives \(\frac{81/6}{9/2} = 3\).
\[\frac{27/2}{9/2} = 3\]
Questions
The graphs show the first six terms of a series. Green bars are positive terms, and red bars are negative terms.
What is \(\left|\frac{a_4}{a_3}\right|\)?
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
- \(-\frac{2}{3}\)
- 3
Answer
\(\frac{2}{3}\)
Between which two neighboring terms does \(\left|\frac{a_{n+1}}{a_n}\right|\) equal 1?
- \(a_1\) and \(a_2\)
- \(a_2\) and \(a_3\)
- \(a_3\) and \(a_4\)
- None of them
Answer
\(a_1\) and \(a_2\)
Which expression is \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(\displaystyle\sum (-1)^n \frac{3^n}{(n+2)!}\), before any simplifying?
- \(\dfrac{3^{n+1}/(n+3)!}{3^n/(n+2)!}\)
- \(\dfrac{3^{n+1}/(n+2)!}{3^n/(n+2)!}\)
- \(\dfrac{3^n/(n+2)!}{3^{n+1}/(n+3)!}\)
- \(-\dfrac{3^{n+1}/(n+3)!}{3^n/(n+2)!}\)
Answer
\(\dfrac{3^{n+1}/(n+3)!}{3^n/(n+2)!}\)
Summary
You write the next term by replacing every n with n + 1, you divide it by the current term, and you take the absolute value.
This works because the ratio compares the lengths of two neighboring terms, and the absolute value removes the signs.
You use it as the first step of the ratio test, before you simplify the ratio and take its limit.
This is one of four skills in topic 10.8 of Which & Why: AP Calculus.