Simplify the ratio
From a long ratio to a short one
The bars show the terms of \(\sum \frac{6^n}{(n+2)!}\), and the dot at each \(n\) shows the ratio of the next term to that term. The fraction above the dots is this ratio, with two powers of 6 and two factorials.
Press "Cancel the powers". Because \(6^{n+1}\) is 6 times \(6^n\), the powers leave a 6. Then press "Cancel the factorials". Because \((n+3)!\) is \(n+3\) times \((n+2)!\), the factorials leave \(n+3\) in the denominator.
The ratio is now \(\frac{6}{n+3}\), and the dots do not move, because cancelling changes the form of the ratio and not its value. Move the slider: the ratio is 1.5 at n = 1, 1 at n = 3, and below 1 after that.
Cancel the common factors
Move the slider to change the base \(c\). The dots move, because the terms change.
Press the buttons to cancel. Each time, a factor that appears on the top and on the bottom cancels, so the fraction becomes shorter.
The dots stay in place, because dividing the top and the bottom by the same nonzero number does not change a fraction.
Notation
Powers with the same base cancel, because \(c^{n+1}\) is \(c^n\) times one more factor \(c\).
\[\frac{c^{n+1}}{c^n} = c\]Factorials cancel when you write the larger one as the smaller one times the missing factors. If the factorial grows by one step, one factor remains, and if it grows by two steps, two factors remain.
\[\begin{aligned}\frac{(n+1)!}{n!} &= n+1\\ \frac{(n+2)!}{n!} &= (n+2)(n+1)\end{aligned}\]A sum in \(n\), such as \(n+1\), does not cancel with \(n\), so it stays as a fraction. To find its limit later, you divide the top and the bottom by the highest power of \(n\).
\[\frac{(n+1)^2}{n^2} = \left(1 + \frac{1}{n}\right)^2\]For the series in the graph, \(\sum \frac{n\,4^n}{(n+1)!}\), the three rules give the simplified ratio.
When it fails
If you cancel the \(n\) in \(\frac{n+1}{n}\) and write 1, the ratio becomes \(\frac{4}{n+2}\). This is wrong, because the \(n\) in the top is added to 1, so it is a term and not a factor.
The red dots show this wrong ratio, which at \(n = 2\) is 1 instead of \(\frac32\). Cancel only factors that multiply the whole top and the whole bottom.
Example 1: A power and a linear factor
Simplify \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(a_n = \dfrac{3n+1}{4^n}\).
Write the ratio with the next term on the top. Replacing \(n\) with \(n+1\) turns \(3n+1\) into \(3n+4\).
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{3n+4}{4^{n+1}} \cdot \frac{4^n}{3n+1}\]Cancel the powers of 4. The power \(4^{n+1}\) is 4 times \(4^n\).
\[\frac{4^n}{4^{n+1}} = \frac{1}{4}\]The sums \(3n+4\) and \(3n+1\) do not cancel, so the simplified ratio is a fraction in \(n\), and the dots show its values.
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{3n+4}{4(3n+1)}\]Test at \(n = 1\). The first two terms are 1 and \(\frac{7}{16}\), so their ratio is \(\frac{7}{16}\), which matches the first dot and the formula.
\[\frac{7}{4 \cdot 4} = \frac{7}{16}\]
Example 2: Factorials of different sizes
Simplify \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(a_n = \dfrac{(2n)!}{5^n \, n!}\).
Write the ratio with the next term on the top.
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{(2n+2)!}{5^{n+1}(n+1)!} \cdot \frac{5^n \, n!}{(2n)!}\]Cancel the powers of 5 and the factorials of n. Each cancellation leaves one factor.
\[\begin{aligned}\frac{5^n}{5^{n+1}} &= \frac{1}{5}\\ \frac{n!}{(n+1)!} &= \frac{1}{n+1}\end{aligned}\]A tempting wrong move: The factorials \((2n+2)!\) and \((2n)!\) leave the factor \(2n+2\).
\[\frac{(2n+2)!}{(2n)!} = 2n+2\]This gives the red dot at \(\frac25\), far below the first dot. At \(n = 1\) the quotient is \(\frac{4!}{2!} = 12\), not \(2n+2 = 4\), because \((2n+2)!\) has two more factors than \((2n)!\).
Write the larger factorial as \((2n+2)! = (2n+2)(2n+1)\,(2n)!\), so two factors remain.
\[\frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)\]Join the parts. The factor \(2n+2\) equals \(2(n+1)\), so the factor \(n+1\) cancels. At \(n = 1\) the ratio is \(\frac65\), which is the first dot.
\[\frac{(2n+2)(2n+1)}{5(n+1)} = \frac{2(2n+1)}{5}\]
Practice
Simplify \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(a_n = \dfrac{3^n (n+1)}{(n+2)!}\).
Cancel the powers of 3 first. Then write \((n+3)!\) as \((n+3)(n+2)!\). Test your result at \(n = 1\) against the first dot.
Solution
Write the ratio with the next term on the top.
\[\frac{3^{n+1}(n+2)/(n+3)!}{3^n(n+1)/(n+2)!}\]Cancel the powers of 3 and the factorials.
\[\begin{aligned}\frac{3^{n+1}}{3^n} &= 3\\ \frac{(n+2)!}{(n+3)!} &= \frac{1}{n+3}\end{aligned}\]Join the parts, where the sums \(n+2\) and \(n+1\) do not cancel. At \(n = 1\) the first two terms are 1 and \(\frac98\), so their ratio is \(\frac98\), as the formula gives.
\[\frac{3(n+2)}{(n+1)(n+3)}\]
Questions
The graphs show the terms of \(\sum \frac{4^n}{(n+1)!}\) as bars. The dots show the ratio of neighboring terms, and the dashed line is at 1.
What is the ratio \(\left|\frac{a_5}{a_4}\right|\)?
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
- 0.71
- 1
Answer
\(\frac{2}{3}\)
For which \(n\) does the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\) equal 1?
- \(n = 2\)
- \(n = 1\)
- \(n = 3\)
- For no n
Answer
\(n = 2\)
Which fraction equals \(\left|\frac{a_{n+1}}{a_n}\right|\) for \(a_n = \dfrac{7^n \, n!}{(3n)!}\)?
- \(\dfrac{7}{3(3n+1)(3n+2)}\)
- \(\dfrac{7(n+1)}{3n+1}\)
- \(\dfrac{7}{(3n+1)(3n+2)}\)
- \(\dfrac{7(3n+1)(3n+2)(3n+3)}{n+1}\)
Answer
\(\dfrac{7}{3(3n+1)(3n+2)}\)
Summary
You cancel powers with the same base, and factorials by writing out the missing factors, while sums in n stay in the fraction.
This works because cancelling a common factor does not change the value of a fraction.
You use it to turn the ratio into a short expression whose limit you can find.
This is one of four skills in topic 10.8 of Which & Why: AP Calculus.