A Mixed Workshop on Ratio, Root, and Test Selection

Choosing a Test from the Shape of a Term

A series can combine several kinds of growth: factorials, exponential factors, polynomial factors, and expressions raised to the index. The central skill in this workshop is not applying both the ratio and root tests automatically. It is recognizing which calculation will reveal the decisive behavior with the least unnecessary algebra.

Factorials often simplify when consecutive terms are divided. Expressions raised to the index often simplify when an \(n\)th root is taken. Polynomial factors may remain in the calculation, but their effect can often be evaluated using familiar limits such as \(n^{1/n}\to 1\). The result must still be interpreted carefully: a limit below one proves absolute convergence, a limit above one proves divergence, and a limit equal to one is inconclusive.

Definition: For a series \(\sum_{n=1}^{\infty} a_n\), the ratio test examines the limit of \(\left|a_{n+1}/a_n\right|\), when the terms are eventually nonzero. The root test examines the limit of \(\sqrt[n]{|a_n|}\). For either test, a limit below one proves absolute convergence, a limit above one proves divergence, and a limit equal to one gives no conclusion.

“Inconclusive” means only that the particular test does not settle the question. It does not mean that the series has no convergence behavior, or that the series is necessarily divergent.

Why the Tests Work

The key idea behind both tests is comparison with a geometric series. A geometric series \(\sum q^n\) converges when \(0\le q<1\), and its terms do not tend to zero when \(q>1\).

Theorem: Suppose \(a_n\) is eventually nonzero. If \(\lim_{n\to\infty}|a_{n+1}/a_n|=L\), or if \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=L\), then \(L<1\) implies that \(\sum a_n\) converges absolutely, and \(L>1\) implies that \(\sum a_n\) diverges. Neither test decides the case \(L=1\).

Here is the justification for the ratio test when \(L<1\). Choose a number \(q\) with \(L<q<1\). By the definition of a limit, for all sufficiently large \(n\), \(|a_{n+1}/a_n|<q\). Repeatedly applying this inequality bounds the terms by a constant times the terms of a convergent geometric series.

|a_{N+k}| \le |a_N|q^k \qquad (k\ge 0).

Thus the tail of \(\sum |a_n|\) is bounded by a constant multiple of \(\sum_{k=0}^{\infty}q^k\), so the original series converges absolutely. If \(L>1\), choose \(q\) with \(1<q<L\). Eventually, \(|a_{n+1}|>q|a_n|\), so the terms cannot tend to zero; the series diverges by the term test.

For the root test, if \(L<1\), choose \(q\) with \(L<q<1\). Eventually \(\sqrt[n]{|a_n|}<q\), so \(|a_n|<q^n\). Comparison with a geometric series proves absolute convergence. If \(L>1\), choose \(q\) with \(1<q<L\). Eventually \(|a_n|>q^n\), so the terms do not tend to zero. When \(L=1\), neither argument supplies a geometric bound in the needed direction.

Worked Examples

Worked Example: A Factorial with Exponential and Polynomial Factors

Classify the series

\sum_{n=1}^{\infty}\frac{n^4 3^n}{2^n n!}.

Write its positive \(n\)th term as \(a_n=n^4 3^n/(2^n n!)\). Since the term contains \(n!\), dividing consecutive terms should simplify the factorial directly:

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n+1)^4 3^{n+1}}{2^{n+1}(n+1)!} \cdot \frac{2^n n!}{n^4 3^n}\\ &= \frac{3}{2}\cdot\frac{(n+1)^4}{n^4}\cdot\frac{1}{n+1}\\ &= \frac{3}{2}\cdot\frac{(n+1)^3}{n^4}. \end{aligned}

Consequently,

\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \frac{3}{2}\lim_{n\to\infty}\frac{(n+1)^3}{n^4} =0.

The ratio limit is below one, so the ratio test proves absolute convergence. The calculation is consistent: the remaining factor has degree three in the numerator and degree four in the denominator, and therefore tends to zero. Since every term is positive, this is also convergence of the given series.

Worked Example: An nth Power with a Polynomial Factor

Classify

\sum_{n=1}^{\infty} n^2 \left(\frac{2n+1}{5n+4}\right)^n.

Let \(a_n=n^2((2n+1)/(5n+4))^n\). The expression in parentheses is raised to the index, so take the \(n\)th root:

\sqrt[n]{a_n} = \sqrt[n]{n^2}\,\frac{2n+1}{5n+4} = n^{2/n}\frac{2n+1}{5n+4}.

Use \(n^{1/n}\to1\), so \(n^{2/n}=(n^{1/n})^2\to1\), and divide the rational factor’s numerator and denominator by \(n\):

\lim_{n\to\infty}\sqrt[n]{a_n} = 1\cdot \lim_{n\to\infty}\frac{2+1/n}{5+4/n} = \frac{2}{5}.

Since \(2/5<1\), the root test proves absolute convergence. The positive terms mean this is convergence of the displayed series itself. In particular, the polynomial factor \(n^2\) contributes a root limit of one; it does not alter the decisive factor \(2/5\).

Worked Example: Factorial Quotients and an Exponential Factor

Classify the series

\sum_{n=1}^{\infty} \frac{2^n n^2(n!)^2}{(3n)!}.

Use the ratio test because the factorial quotient is especially simple across consecutive indices. Set \(a_n=2^n n^2(n!)^2/(3n)!\). Then

\begin{aligned} \frac{a_{n+1}}{a_n} &= 2\cdot\frac{(n+1)^2}{n^2} \cdot\frac{((n+1)!)^2}{(n!)^2} \cdot\frac{(3n)!}{(3n+3)!}\\ &= 2\cdot\frac{(n+1)^2}{n^2} \cdot(n+1)^2 \cdot\frac{1}{(3n+1)(3n+2)(3n+3)}\\ &= \frac{2(n+1)^4}{n^2(3n+1)(3n+2)(3n+3)}. \end{aligned}

The numerator has degree four in \(n\), while the denominator has degree five. More explicitly, dividing numerator and denominator by \(n^5\) gives a numerator tending to zero and a denominator tending to \(27\):

\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{2(1+1/n)^4} {(3+1/n)(3+2/n)(3+3/n)n} =0.

The ratio test proves absolute convergence, and the terms are positive, so the series converges. The final limit agrees with the degree comparison: the denominator has one more factor of order \(n\) than the numerator.

The Boundary Value One

A result of one is not a weak version of convergence or divergence. It is a genuine boundary at which the ratio and root tests may fail to distinguish very different series. The following pair makes that limitation explicit.

Worked Example: Two Series the Tests Cannot Decide

For \(a_n=1/n\), the ratio and root limits are

\frac{a_{n+1}}{a_n}=\frac{n}{n+1}\longrightarrow1, \qquad \sqrt[n]{a_n}=n^{-1/n}\longrightarrow1.

Both tests are inconclusive. Nevertheless, the harmonic series diverges. In the block of indices \(2^k\le n<2^{k+1}\), there are \(2^k\) terms and each satisfies \(1/n\ge 1/2^{k+1}\). Therefore each block has sum at least

2^k\cdot\frac{1}{2^{k+1}}=\frac12.

Infinitely many blocks each contribute at least \(1/2\), so the partial sums are unbounded and the series diverges.

Now take \(b_n=1/n^2\). Again,

\frac{b_{n+1}}{b_n} = \left(\frac{n}{n+1}\right)^2\longrightarrow1, \qquad \sqrt[n]{b_n}=n^{-2/n}\longrightarrow1.

Again, both tests are inconclusive. But on the same block \(2^k\le n<2^{k+1}\), there are \(2^k\) terms and \(1/n^2\le1/2^{2k}\). Thus the block sum is at most

2^k\cdot\frac{1}{2^{2k}}=\frac{1}{2^k}.

The sum of these upper bounds over \(k\ge0\) is a convergent geometric series. The first block consists just of \(n=1\), and the bounds for all the blocks show that \(\sum 1/n^2\) converges. The two examples have identical ratio and root test outcomes, yet one diverges and the other converges. The calculations are checked by the block estimates, not by the inconclusive test limits.

A Practical Selection Process

Before calculating, look for the part of the term that will simplify most decisively. A factorial is not automatically a reason to use the ratio test, nor is an \(n\)th power automatically a reason to use the root test. The useful question is which operation reduces the term to a limit that can be evaluated clearly.

Check the term’s structure.
Notice factorials, exponential factors, polynomial factors, and quantities raised to \(n\). Confirm that the terms are eventually nonzero if using the ratio test.

Choose a simplifying operation.
Try a consecutive-term ratio when factorials cancel cleanly. Try an \(n\)th root when the term contains a substantial expression raised to \(n\).

Evaluate the limit before concluding.
Keep polynomial factors visible, simplify the exact expression, and then compute the limit. Do not infer convergence merely because a term looks small.

Match the conclusion to the limit.
A limit below one proves absolute convergence; a limit above one proves divergence; a limit equal to one is inconclusive. If inconclusive, use a different valid argument.

A Common Mistake: Treating One as a Verdict

A frequent error is to calculate a ratio limit of one and then declare the series divergent because the test “does not prove convergence.” That confuses failure to prove convergence with proof of divergence. For example, with \(a_n=1/n^2\), the calculation

\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{n^2}{(n+1)^2} =1

does not prove divergence. It proves only that the ratio test gives no conclusion. The block estimate above establishes that \(\sum 1/n^2\) converges. The correction is to state the test’s result precisely, then seek another argument if the test is inconclusive.

The opposite error is also possible: treating a ratio or root limit equal to one as evidence that the series converges. The harmonic series has the same boundary limit and diverges. A boundary result is not evidence either way; it is a signal to change methods.

Why Test Selection Matters

Choosing a useful test reduces algebra and makes the logic of the conclusion easier to check. In the factorial examples, consecutive terms exposed the factorial cancellation. In the nth-power example, taking roots removed the exponent \(n\) and left a simple limiting ratio. These choices are not mechanical rules, but they are reliable starting points.

Always separate three tasks: simplify correctly, calculate the limit correctly, and apply the criterion correctly. A correct calculation with an incorrect interpretation is still an incorrect classification. In particular, “less than one,” “greater than one,” and “equal to one” lead to three different statements about what the test establishes.

Check Your Understanding

For each question, identify what the chosen test establishes, rather than guessing from the appearance of the terms.

For \(a_n=5^n/n!\), what ratio limit do you obtain, and what does it imply for \(\sum a_n\)?
For \(a_n=n^3(1/4)^n\), what is the root-test limit, and what conclusion follows?
If the ratio limit is \(3/2\), what does the ratio test establish?
If the root-test limit is one, what conclusion may be drawn from the root test alone?
Why do \(\sum 1/n\) and \(\sum 1/n^2\) show that a limit of one is inconclusive?