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Ratio, Root, and Test Selection · Tutorial 689 of 998

The Ratio Test with Limit Superior

Learn how the limit superior of successive-term ratios gives a rigorous convergence test and why exceeding one alone does not prove divergence.

Advanced 9 min read

What You'll Learn

  • Define the limit superior and limit inferior of successive-term ratios.
  • Prove absolute convergence when the limit superior is below one.
  • Identify why a limit superior above one does not by itself prove divergence.
  • Use the limit inferior criterion to establish divergence.
  • Compare examples where ratios fluctuate but the series converges or diverges.

Why Use a Limit Superior?

The ratio test studies how the size of one term compares with the size of the next. When the successive-term ratios approach a single limit, the test gives a direct conclusion. But those ratios need not approach a limit: they may fluctuate indefinitely.

The limit superior provides a way to describe the eventual upper behavior of a fluctuating sequence. It gives a useful convergence criterion for series. There is an important asymmetry, however: a limit superior above one does not by itself prove divergence. The correct divergence condition uses the limit inferior, or a stronger assumption that the ratios eventually stay above one.

Key distinction: If the limit superior of the absolute successive-term ratios is below one, the series converges absolutely. If that limit superior is above one, the ratio test is inconclusive unless additional information is available.

Definitions

Let \(\sum_{n=1}^{\infty} a_n\) be a series whose terms are eventually nonzero. Define its absolute successive-term ratios by

$$ r_n=\frac{|a_{n+1}|}{|a_n|}. $$

For a sequence of nonnegative numbers \(r_n\), its limit superior is the limit of the suprema of its tails:

$$ \limsup_{n\to\infty} r_n = \lim_{N\to\infty}\sup\{r_n:n\ge N\}. $$

The tail suprema decrease or stay the same as \(N\) increases, so their limit exists in the extended nonnegative real numbers. Informally, the limit superior records the largest values that continue to occur arbitrarily far out in the sequence. The limit inferior is defined using tail infima:

$$ \liminf_{n\to\infty} r_n = \lim_{N\to\infty}\inf\{r_n:n\ge N\}. $$

It records the smallest values that persist arbitrarily far out. If the ratios have an ordinary limit, their limit superior and limit inferior equal that limit. When ratios fluctuate, the two quantities can differ.

The Convergence Criterion

Theorem: Suppose \(a_n\) is nonzero for all sufficiently large \(n\), and let \(L=\limsup_{n\to\infty}|a_{n+1}|/|a_n|\). If \(L<1\), then \(\sum_{n=1}^{\infty}a_n\) converges absolutely. If \(L>1\), this condition alone gives no conclusion about convergence or divergence.

The first conclusion follows because, eventually, every ratio is bounded above by a fixed number less than one. The terms in the tail are then bounded by a geometric sequence.

1
Choose a geometric ratio.
Since \(L<1\), choose \(q\) with \(L<q<1\).
2
Bound all sufficiently late ratios.
By the definition of limit superior, \(r_n<q\) for all sufficiently large \(n\).
3
Compare with a geometric series.
The tail terms satisfy \(|a_{N+k}|\le |a_N|q^k\) for a sufficiently large \(N\) and every integer \(k\ge 0\).
4
Conclude absolute convergence.
The geometric series \(\sum_{k=0}^{\infty}|a_N|q^k\) converges, so comparison shows that \(\sum |a_n|\) converges.

Why the Convergence Proof Works

Here is the complete comparison argument. Choose \(q\) such that \(L<q<1\). If ratios were at least \(q\) arbitrarily far out, every sufficiently late tail would have supremum at least \(q\), contradicting \(L<q\). Thus there is an index \(N\) such that \(r_n<q\) for every \(n\ge N\).

Starting with \(|a_N|\), apply this bound repeatedly:

$$ |a_{N+1}|<q|a_N|,\qquad |a_{N+2}|<q|a_{N+1}|<q^2|a_N|, \qquad |a_{N+k}|\le q^k|a_N|. $$

Consequently, for every positive integer \(m\),

$$ \sum_{k=0}^{m}|a_{N+k}| \le |a_N|\sum_{k=0}^{m}q^k \le \frac{|a_N|}{1-q}. $$

The partial sums of the nonnegative tail \(\sum_{k=0}^{\infty}|a_{N+k}|\) are therefore bounded above, so that tail converges. Adding the finitely many terms before \(a_N\) does not affect convergence. Hence \(\sum |a_n|\) converges, which is precisely absolute convergence of \(\sum a_n\).

Worked Example: A Limit Superior Below One

Worked Example: A Limit Superior Below One

Consider \(a_n=n/3^n\) for \(n\ge 1\). Compute the successive-term ratios and use the limit-superior criterion.

First calculate:

$$ r_n = \frac{a_{n+1}}{a_n} = \frac{(n+1)/3^{n+1}}{n/3^n} = \frac{n+1}{3n} = \frac{1+1/n}{3}. $$

As \(n\) tends to infinity, \(1/n\) tends to zero, so \(r_n\) tends to \(1/3\). Therefore,

$$ \limsup_{n\to\infty}r_n=\frac13<1. $$

For a direct geometric bound, when \(n\ge2\),

$$ \frac{n+1}{3n}\le\frac12 \quad\Longleftrightarrow\quad 2n+2\le3n \quad\Longleftrightarrow\quad n\ge2. $$

Thus \(a_{n+1}\le a_n/2\) for every \(n\ge2\), and repeated application gives \(a_{2+k}\le a_2(1/2)^k\). Since \(a_2=2/9\), the tail is bounded by a convergent geometric series. The series \(\sum n/3^n\) therefore converges absolutely. Checking the ratio calculation, substituting the original terms gives \(((n+1)/3^{n+1})/(n/3^n)=(n+1)/(3n)\), as required.

Why a Limit Superior Above One Is Not Enough

A limit superior above one means that ratios greater than one occur arbitrarily far out. It does not say that all sufficiently late ratios are greater than one. Ratios that are sometimes large and sometimes small can still produce terms that shrink overall.

Worked Example: Ratios Above One Can Still Give a Convergent Series

Define positive terms recursively by \(a_1=1\), with \(a_{n+1}/a_n=2\) when \(n\) is odd and \(a_{n+1}/a_n=1/4\) when \(n\) is even.

The first terms confirm the pattern:

$$ a_1=1,\qquad a_2=2,\qquad a_3=\frac12,\qquad a_4=1,\qquad a_5=\frac14. $$

Each pair of ratios multiplies the term by \(2(1/4)=1/2\). More explicitly, for \(k\ge1\),

$$ a_{2k-1}=\left(\frac12\right)^{k-1}, \qquad a_{2k}=2\left(\frac12\right)^{k-1}. $$

The sum of each pair is \(3(1/2)^{k-1}\), so grouping consecutive positive terms gives

$$ \sum_{n=1}^{\infty}a_n = \sum_{k=1}^{\infty}(a_{2k-1}+a_{2k}) = \sum_{k=1}^{\infty}3\left(\frac12\right)^{k-1} = 6. $$

The ratios alternate between \(2\) and \(1/4\), so their limit superior is \(2\), which is greater than one. Nevertheless, the series converges. The calculation checks both features: the ratio pattern is as defined, and the paired geometric sum is finite. Thus a limit superior above one cannot, by itself, prove divergence.

The Correct Ratio Condition for Divergence

To force divergence by term size, the ratios must stay above one eventually, not merely exceed one repeatedly. A convenient sufficient condition expresses this using the limit inferior.

Theorem: If \(a_n\) is eventually nonzero and \(\liminf_{n\to\infty}|a_{n+1}|/|a_n|>1\), then \(a_n\) does not tend to zero. Therefore \(\sum a_n\) diverges by the term test.

To prove this, choose \(q\) strictly between \(1\) and the limit inferior. The definition of limit inferior implies that \(r_n>q\) for every sufficiently large \(n\). For some such index \(N\), repeated multiplication gives

$$ |a_{N+k}|>|a_N|q^k \qquad(k\ge0). $$

Because \(q>1\) and \(|a_N|>0\), the right-hand side does not tend to zero; it grows without bound. Hence the terms \(a_n\) do not tend to zero. A necessary condition for a series to converge is that its terms tend to zero, so the series diverges.

Worked Example: Ratios Eventually Above One

Let \(a_1=1\), and define positive terms by \(a_{n+1}/a_n=3\) for odd \(n\) and \(a_{n+1}/a_n=2\) for even \(n\). Every ratio is at least \(2\), so the limit inferior is \(2>1\).

For every \(n\), \(a_{n+1}\ge2a_n\). Starting from \(a_1=1\), induction gives

$$ a_n\ge2^{n-1}. $$

Indeed, this holds at \(n=1\); if \(a_n\ge2^{n-1}\), then \(a_{n+1}\ge2a_n\ge2^n\), which establishes the next case. In particular, the terms do not tend to zero, so the series diverges. The ratio check is exact: the ratios are \(3,2,3,2,\ldots\), whose limit inferior is \(2\), and the lower bound verifies the term-test conclusion.

A Common Mistake

A tempting but incorrect argument is: “The limit superior of the ratios is greater than one, so the terms eventually grow and the series diverges.” The conclusion does not follow. A limit superior above one guarantees only that ratios greater than one recur arbitrarily late; it does not establish that every late ratio exceeds one. The convergent example above has limit superior \(2\) but repeatedly has a ratio of \(1/4\).

The correction is to distinguish the two situations. Use \(\limsup r_n<1\) to prove convergence. To prove divergence through term growth, establish \(\liminf r_n>1\), or verify directly that \(r_n\) is bounded below by some fixed number greater than one for all sufficiently large \(n\). If the limit superior is above one but the limit inferior is not, the ratio information alone may be inconclusive.

Likewise, when the limit superior equals one, this criterion gives no conclusion. A test’s boundary value is not a verdict: the series may converge or diverge, and further information is needed.

When This Form of the Test Is Useful

The ordinary ratio test is especially straightforward when successive-term ratios have a limit. The limit-superior form extends its convergence conclusion to sequences whose ratios fluctuate, as long as their eventual upper behavior remains strictly below one. The distinction between upper and lower behavior is essential: a persistent upper bound below one gives geometric decay, while divergence by ratio growth requires a persistent lower bound above one.

When applying the test, first form the ratio using absolute values, simplify it carefully, and then identify which eventual bound the calculation actually establishes. Do not replace “infinitely often above one” with “eventually above one.” That difference determines whether the ratio information proves divergence or leaves the question open.

Check Your Understanding

For each question, distinguish carefully between an eventual bound and a bound that occurs only along a subsequence.

  1. What does \(\limsup r_n<1\) imply about all sufficiently late ratios?
  2. Why does \(\limsup r_n>1\) not by itself prove that the terms fail to tend to zero?
  3. If \(\liminf r_n>1\), what inequality can be established for every sufficiently late ratio?
  4. A series has positive terms and ratios alternating between \(3/2\) and \(1/5\). Does the limit superior alone settle convergence? Explain.
  5. What conclusion follows from the ratio test if the limit superior equals one?

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