Choosing a Test by Predicting the Algebra
A convergence test is useful only if it leads to a conclusion, and some tests make that conclusion much easier to reach than others. When considering the ratio test and the root test, the first question is not merely whether either test can be applied. Ask instead: which expression is likely to simplify most?
The ratio test compares consecutive terms, so it is often a natural choice when factorials, products, or other consecutive-index expressions cancel cleanly. The root test takes the \(n\)th root of the \(n\)th term, so it is often a natural choice when the term is visibly an \(n\)th power. These are useful predictions, not rigid rules: sometimes both tests are equally simple, and sometimes both return an inconclusive limit.
What Each Test Examines
Let \(\sum_{n=1}^{\infty} a_n\) be a series. The ratio test examines the limit of the absolute value of the quotient of consecutive terms. If \(a_n\neq 0\) for all sufficiently large \(n\) and
exists, then the series converges absolutely if \(L<1\), and diverges if \(L>1\). If \(L=1\), the test gives no conclusion.
The root test examines the \(n\)th root of the absolute value of each term. If
exists, then the series converges absolutely if \(L<1\), and diverges if \(L>1\). If \(L=1\), the test gives no conclusion. In either test, the limit must be calculated correctly; a familiar-looking expression is not enough to guarantee a useful result.
The tests have matching decision thresholds, but they inspect different features of the terms. The ratio test can expose cancellation between \(a_{n+1}\) and \(a_n\). The root test can remove an exponent \(n\) that applies to the whole term.
Why These Forms Suggest Different Tests
Suppose a term has the form \(a_n=b_n^n\), where \(b_n\geq 0\). Taking the \(n\)th root gives \(\sqrt[n]{a_n}=b_n\). The exponent that may have made the term look complicated disappears exactly. This is why an explicit \(n\)th power is a strong clue to try the root test.
For a factorial term, the quotient may instead cancel nearly all of the factorial structure. For example, \((n+1)!/n!=n+1\). This cancellation is why factorials and products indexed by consecutive integers often point toward the ratio test.
Look for factorials or consecutive products, an explicit \(n\)th power, or both.
Ask what cancels in \(\left|a_{n+1}/a_n\right|\) and what remains in \(\sqrt[n]{|a_n|}\).
Calculate that limit, then apply the relevant threshold. A limit of one is inconclusive, even if the algebra was easy.
Worked Example: Factorials Favor the Ratio Test
Worked Example: Factorials Favor the Ratio Test
Determine whether \(\sum_{n=1}^{\infty}\frac{3^n}{n!}\) converges. The factorial suggests testing consecutive terms: the factorial in \(a_{n+1}\) should cancel with the one in \(a_n\).
Here \(a_n=\frac{3^n}{n!}\), and every term is positive. The quotient is
Since \(\lim_{n\to\infty}\frac{3}{n+1}=0\), the ratio-test limit is \(L=0<1\). Thus the series converges absolutely. Because its terms are positive, this is also ordinary convergence.
Check: the quotient has the expected cancellation, and its limit lies strictly below one. That matches the ratio test's convergence condition.
Worked Example: An Explicit nth Power Favors the Root Test
Worked Example: An Explicit nth Power Favors the Root Test
Determine whether \(\sum_{n=1}^{\infty}\left(\frac{3n+1}{8n+7}\right)^n\) converges. The term is an \(n\)th power, so taking the \(n\)th root removes that power directly.
Set \(a_n=\left(\frac{3n+1}{8n+7}\right)^n\). The base is positive for \(n\geq1\), and therefore
As \(n\to\infty\), the last expression tends to \(\frac{3}{8}\). Thus the root-test limit is \(L=\frac{3}{8}<1\), and the series converges absolutely.
Check: taking the root has eliminated the exponent \(n\), leaving a rational expression whose limit is below one. The root-test conclusion therefore applies.
Worked Example: A Root Can Simplify a Less Obvious Power
Worked Example: A Root Can Simplify a Less Obvious Power
Consider \(\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}\). The exponent is \(n^2\), rather than just \(n\), but taking the \(n\)th root still reduces it to a manageable power. This makes the root test a reasonable first choice.
Write \(a_n=\left(\frac{n}{n+1}\right)^{n^2}\). Since \(a_n>0\),
Using the standard limit \(\lim_{n\to\infty}(1+\frac{1}{n})^n=e\), the root-test limit is \(L=e^{-1}\). Since \(e>1\), \(e^{-1}<1\), so the series converges absolutely.
Check: the \(n\)th root changes the exponent \(n^2\) to \(n\), and the resulting limit is strictly less than one. That verifies the required condition.
When Both Tests Look Reasonable
The choice is not always exclusive. Consider \(\sum_{n=1}^{\infty}\frac{n^2}{4^n}\). The ratio test is especially direct because the polynomial terms at consecutive indices form a simple quotient:
Since \(\frac{1}{4}<1\), the series converges absolutely. The root test also works: \(\sqrt[n]{a_n}=\frac{n^{2/n}}{4}\to\frac{1}{4}\), using \(n^{1/n}\to1\). In this case, either test gives the same conclusion. A good selection is about choosing an efficient route, not proving that the other test is unusable.
Sometimes both expressions are simple but neither test decides the series. For \(\sum_{n=1}^{\infty}\frac{1}{n^2}\), the ratio-test limit is \(\lim_{n\to\infty}(\frac{n}{n+1})^2=1\), and the root-test limit is \(\lim_{n\to\infty}n^{-2/n}=1\). Each test is inconclusive. This does not mean the series diverges or that the calculations failed; it means these tests do not settle its behavior.
Common Mistake: Dropping a Changing Exponent
A tempting but incorrect shortcut is to form the ratio of the bases when the terms themselves have changing exponents. For the terms in the second worked example, the first two terms are \(a_1=(4/15)^1\) and \(a_2=(7/23)^2\). Their actual quotient is
Dropping the exponents would instead give \(\frac{7/23}{4/15}=\frac{105}{92}\), which is not the quotient of the terms. In general, for \(a_n=(\frac{3n+1}{8n+7})^n\), the ratio is
This quotient is valid, but it is less transparent than the \(n\)th root. The correction is not that the ratio test is forbidden; it is to preserve all changing exponents and choose the root test when it reveals the simpler expression.
A Practical Decision
| Feature of the term | Likely first choice | Reason to consider it |
|---|---|---|
| Factorials or consecutive products | Ratio test | Neighboring terms may cancel many factors. |
| A term of the form \(b_n^n\) or a related nth power | Root test | The nth root may remove the main exponent. |
| Both forms simplify cleanly | Either test | Compare the resulting limits and use a decisive one. |
| Either limit equals one | Neither test decides | A different argument or test is needed. |
This choice matters because unnecessary algebra creates opportunities for errors, especially when indices and exponents change at the same time. Predicting the simplification first helps keep the calculation focused. It also prevents a mistaken assumption that a test must work simply because it is applicable. The decisive question is always the value of the resulting limit and whether it falls on the convergent or divergent side of the threshold.
Check Your Understanding
For each question, focus first on what the proposed test is expected to simplify.
- For \(\sum_{n=1}^{\infty}\frac{5^n}{n!}\), which test would you try first, and what feature suggests that choice?
- For \(\sum_{n=1}^{\infty}\left(\frac{2n+1}{7n+3}\right)^n\), what does the root-test expression become, and what is its limit?
- State the conclusions of the ratio and root tests when their limits are less than one, greater than one, or equal to one.
- Why is it incorrect to form the ratio of the bases alone for terms whose exponents change with \(n\)?
- If both tests give a limit of one, what can you conclude from those tests?