AP Calculus BC: Ratio Test

Determining whether an infinite series converges or diverges is a central challenge in calculus. Among the many tests available, the Ratio Test is a powerful and frequently used tool, particularly when series terms involve factorials, exponentials, or products that grow rapidly. Its power lies in a straightforward calculation: it examines the limit of the absolute value of the ratio of consecutive terms. Mastering this test not only streamlines your problem-solving but also deepens your understanding of how series behave as their terms evolve.

The Formal Statement of the Ratio Test

Let ${\sum }_{n=1}^{\infty }{a}_n$ be an infinite series with nonzero terms. The Ratio Test instructs you to compute the limit of the absolute value of the ratio of consecutive terms:

$$L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$$

The convergence of the series is then determined by the value of this limit $L$:

• If $L<1$, the series converges absolutely (and therefore converges).
• If $L>1$ (including $L=\infty$), the series diverges.
• If $L=1$, the Ratio Test is inconclusive. The series may converge conditionally, converge absolutely, or diverge.

This test is a form of comparison to a geometric series. If the limit $L$ is less than 1, the terms are eventually shrinking by a factor less than 1, similar to a convergent geometric series $a{r}^n$. If $L>1$, the terms are eventually growing, preventing the sum from settling to a finite limit.

Applying the Ratio Test: A Step-by-Step Process

Applying the test is a methodical procedure. Let's demonstrate with a classic example: ${\sum }_{n=1}^{\infty }\frac{n^5}{5^n}$.

1. Identify $a_n$ and $a_{n+1}$. Here, $a_n=\frac{n^5}{5^n}$. To find $a_{n+1}$, replace every $n$ with $(n+1)$: $a_{n+1}=\frac{(n+1{)}^5}{5^{n+1}}$.

1. Construct the absolute ratio $\left|\frac{a_{n+1}}{a_n}\right|$. Since both terms are positive for $n\ge 1$, the absolute value bars are often handled implicitly.

$$\frac{a_{n+1}}{a_n}=\frac{(n+1{)}^5/5^{n+1}}{n^5/5^n}=\frac{(n+1{)}^5}{5^{n+1}}\cdot \frac{5^n}{n^5}=\frac{(n+1{)}^5}{n^5}\cdot \frac{1}{5}$$

1. Simplify the expression. We can rewrite the ratio of powers:

$$\frac{a_{n+1}}{a_n}={\left(\frac{n+1}{n}\right)}^5\cdot \frac{1}{5}={\left(1+\frac{1}{n}\right)}^5\cdot \frac{1}{5}$$

1. Take the limit as $n\to \infty$.

$$L=\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^5\cdot \frac{1}{5}=(1{)}^5\cdot \frac{1}{5}=\frac{1}{5}$$

1. State the conclusion. Since $L=\frac{1}{5}<1$, by the Ratio Test, the series ${\sum }_{n=1}^{\infty }\frac{n^5}{5^n}$ converges absolutely.

This process showcases a common pattern: when $a_n$ involves both a polynomial factor (like $n^5$) and an exponential factor (like $5^n$), the exponential growth in the denominator will typically dominate, leading to a limit less than 1 and thus convergence.

Special Considerations: Factorials, Exponentials, and Polynomials

The Ratio Test is especially useful for series involving factorials, as they simplify elegantly when taking the ratio. Consider the series ${\sum }_{n=1}^{\infty }\frac{10^n}{n!}$.

Here, $a_n=\frac{10^n}{n!}$ and $a_{n+1}=\frac{10^{n+1}}{(n+1)!}$. Constructing the ratio: $$\frac{a_{n+1}}{a_n}=\frac{10^{n+1}}{(n+1)!}\cdot \frac{n!}{10^n}=\frac{10^{n+1}}{10^n}\cdot \frac{n!}{(n+1)!}=10\cdot \frac{1}{n+1}$$ Crucially, we use the property $(n+1)!=(n+1)\cdot n!$ to cancel the $n!$. Now take the limit: $$L=\underset{n\to \infty }{\lim }\left|\frac{10}{n+1}\right|=0$$ Since $0<1$, the series converges. This result is intuitive: factorial growth in the denominator ($n!$) is vastly faster than exponential growth in the numerator ($10^n$), causing terms to shrink to zero extremely quickly.

A useful mental hierarchy for growth rates, from slowest to fastest, is: logarithmic $\ll$ polynomial $\ll$ exponential $\ll$ factorial. The Ratio Test effectively compares the growth rates of components in $a_{n+1}$ versus $a_n$.

When the Test is Inconclusive (L = 1)

A common hurdle is reaching $L=1$. In this case, the Ratio Test provides no information, and you must use another convergence test. This frequently happens with series where terms behave like $\frac{1}{n^p}$ (a p-series).

For example, apply the Ratio Test to the harmonic series $\sum \frac{1}{n}$: $$L=\underset{n\to \infty }{\lim }\left|\frac{1/(n+1)}{1/n}\right|=\underset{n\to \infty }{\lim }\frac{n}{n+1}=1$$ The test is inconclusive. We know from the p-series test that $\sum \frac{1}{n}$ diverges. Conversely, for $\sum \frac{1}{n^2}$: $$L=\underset{n\to \infty }{\lim }\left|\frac{1/(n+1{)}^2}{1/n^2}\right|=\underset{n\to \infty }{\lim }{\left(\frac{n}{n+1}\right)}^2=1$$ Again, the test is inconclusive, yet this series converges. When you get $L=1$, you should immediately pivot to other tests like the p-series test, comparison test, integral test, or alternating series test.

Common Pitfalls

1. Forgetting the Absolute Value: The formal test requires $\lim \left|\frac{a_{n+1}}{a_n}\right|$. Omitting the absolute value can lead to an incorrect $L$ if the series has alternating signs. Always compute the ratio using the absolute values of the terms, or ensure the ratio itself is positive before taking the limit.

1. Misapplying the Test to the Wrong Sequence: The Ratio Test is for infinite series, not sequences. Do not try to use it to find the limit of a sequence like $a_n=\frac{n!}{10^n}$. It is used to determine the convergence of the sum of such terms, $\sum a_n$.

1. Mishandling Limits in the L=1 Case: When you find $L=1$, the correct conclusion is "The Ratio Test is inconclusive," not "The series converges" or "The series diverges." This is a signal to employ a different test, not to guess based on the borderline result.

1. Algebraic Errors with Factorials and Exponents: The most common computational mistakes occur when simplifying $a_{n+1}/a_n$. Remember key rules: $(n+1)!=(n+1)\cdot n!$, $x^{n+1}=x\cdot x^n$, and $(n+1{)}^p\ne n^p+1$. Write out the simplification step-by-step to avoid errors.

Summary

• The Ratio Test analyzes the limit $L={\lim }_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|$ to determine the convergence of an infinite series.
• If $L<1$, the series converges absolutely. If $L>1$, it diverges. If $L=1$, the test fails and you must use another method.
• This test is particularly effective for series whose terms include factorials, exponential expressions (like $c^n$), or products where growth rates can be easily compared through cancellation in the ratio.
• When applying the test, meticulous algebraic simplification of the ratio $\frac{a_{n+1}}{a_n}$ is crucial, especially when dealing with factorials and compound terms.
• The inconclusive case ($L=1$) is common for series resembling p-series or other slowly diverging/converging series, acting as a trigger to deploy your other convergence test tools.