The ratio test

Theorem

The ratio test

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Statement

Let \( (a_n)_{n \in \mathbb{N}} \) be a sequence of real or complex numbers, and suppose there exists \( N_0 \in \mathbb{N} \) such that \( a_n \neq 0 \) for all \( n \geq N_0 \), so that the ratios \( \left| \frac{a_{n+1}}{a_n} \right| \) are defined from \( N_0 \) onwards. Suppose further that the limit \[ L \;=\; \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] exists in \( [0, \infty) \), or equals \( +\infty \). Then: (i) if \( L \lt 1 \), the series \( \sum_{n} a_n \) converges absolutely, i.e. \( \sum_n |a_n| \lt \infty \); (ii) if \( L \gt 1 \) (including \( L = +\infty \)), the series \( \sum_n a_n \) diverges, and indeed \( a_n \not\to 0 \); (iii) if \( L = 1 \), the test gives no information: there are both convergent and divergent series with \( L = 1 \).

Why it matters

The ratio test is the workhorse convergence test of first-year analysis because it converts a question about an infinite sum into a single limit computation, and that limit is often trivial to evaluate when the terms involve factorials, powers, or products — precisely the terms that appear in power series. It is the standard tool for proving that \( \exp \), \( \sin \), \( \cos \) and the binomial series converge, and (in its \( \limsup \) refinement) it underlies the computation of the radius of convergence of a power series.

Conceptually, the theorem says something sharper than its statement: every series whose term ratios settle below \( 1 \) is eventually dominated by a geometric series. The ratio test is not a new convergence phenomenon — it is the comparison test against \( \sum r^n \), packaged so that the comparison constant never needs to be found explicitly.

Hypotheses

Eventually nonzero terms (\( a_n \neq 0 \) for \( n \geq N_0 \)).If \( a_n = 0 \) for infinitely many \( n \), the ratio \( |a_{n+1}/a_n| \) is undefined infinitely often and the limit \( L \) does not exist as stated. E.g. \( a_n = 0 \) for odd \( n \), \( a_n = 2^{-n} \) for even \( n \): the series converges absolutely, but no ratio-test hypothesis can even be formulated along the full sequence.

The limit \( L = \lim_n |a_{n+1}/a_n| \) exists.Take \( a_n = 2^{-n + (-1)^n} \). The consecutive ratios alternate between \( 2 \) and \( \tfrac{1}{8} \), so no limit exists; \( \limsup = 2 \gt 1 \) while \( \liminf = \tfrac18 \lt 1 \), so even the refined \( \limsup \)/\( \liminf \) forms are silent — yet the series converges (compare \( |a_n| \leq 2 \cdot 2^{-n} \)). If the limit hypothesis is weakened, the correct refinement is: \( \limsup_n |a_{n+1}/a_n| \lt 1 \) forces absolute convergence, and \( \liminf_n |a_{n+1}/a_n| \gt 1 \) forces divergence; nothing follows when \( 1 \) lies between them.

Strict inequality \( L \lt 1 \).Both \( \sum 1/n \) (divergent) and \( \sum 1/n^2 \) (convergent) have \( L = 1 \): the boundary case carries no information. Note also that \( |a_{n+1}/a_n| \lt 1 \) for every \( n \) is not enough — \( \sum 1/n \) satisfies \( \frac{n}{n+1} \lt 1 \) for all \( n \) and diverges. It is the limit being strictly below \( 1 \), leaving a uniform geometric gap, that does the work.

Proof

We prove (i), then (ii); (iii) is witnessed by the pair \( \sum 1/n \), \( \sum 1/n^2 \) exhibited in step 8.

Assume \( L \lt 1 \). Set \( r := \dfrac{L+1}{2} \), so that \( L \lt r \lt 1 \).

Arithmetic: the midpoint of \( [L, 1] \) lies strictly between \( L \) and \( 1 \) because \( L \lt 1 \). A

There exists \( N \geq N_0 \) such that \( \left| \dfrac{a_{n+1}}{a_n} \right| \lt r \) for all \( n \geq N \).

Definition of \( \lim_{n\to\infty} |a_{n+1}/a_n| = L \) applied with \( \varepsilon := r - L \gt 0 \): eventually \( \bigl| |a_{n+1}/a_n| - L \bigr| \lt \varepsilon \), hence \( |a_{n+1}/a_n| \lt L + \varepsilon = r \). A

For every \( k \geq 0 \): \( |a_{N+k}| \leq |a_N| \, r^k \).

Induction on \( k \). Base \( k = 0 \) is trivial. Inductive step: if \( |a_{N+k}| \leq |a_N| r^k \), then by step 2 (with \( n = N+k \geq N \)), \( |a_{N+k+1}| \lt r\,|a_{N+k}| \leq |a_N| r^{k+1} \). This is the key idea: the ratio bound telescopes into geometric decay. B

For every \( M \gt N \): \( \displaystyle\sum_{n=N}^{M} |a_n| \;\leq\; |a_N| \sum_{k=0}^{M-N} r^k \;=\; |a_N|\,\frac{1 - r^{M-N+1}}{1-r} \;\leq\; \frac{|a_N|}{1-r}. \)

Termwise application of step 3, then the finite geometric sum formula; the final bound uses \( 0 \leq r^{M-N+1} \leq 1 \), valid since \( 0 \leq r \lt 1 \). A

The partial sums \( S_M := \displaystyle\sum_{n=0}^{M} |a_n| \) form a nondecreasing sequence bounded above by \( \displaystyle\sum_{n=0}^{N-1} |a_n| + \frac{|a_N|}{1-r} \), hence \( (S_M) \) converges. That is, \( \sum_n |a_n| \lt \infty \).

Nondecreasing because \( |a_n| \geq 0 \); bounded by splitting the sum at \( N \) and applying step 4 to the tail. Convergence is the Monotone Convergence Theorem for real sequences, which is where completeness of \( \mathbb{R} \) (the supremum axiom) enters — the theorem is false over \( \mathbb{Q} \). C

Therefore \( \sum_n a_n \) converges absolutely, and in particular converges.

Step 5 is the definition of absolute convergence. That absolute convergence implies convergence is the standard lemma "absolutely convergent \( \Rightarrow \) convergent", proved from the Cauchy criterion for series together with the triangle inequality \( \left| \sum_{n=p}^{q} a_n \right| \leq \sum_{n=p}^{q} |a_n| \); the Cauchy criterion again rests on completeness. This proves (i). A

Now assume \( L \gt 1 \). There exists \( N \geq N_0 \) with \( |a_{n+1}| \gt |a_n| \) for all \( n \geq N \); hence \( |a_n| \geq |a_N| \gt 0 \) for all \( n \geq N \), so \( a_n \not\to 0 \).

If \( L \) is finite, apply the definition of the limit with \( \varepsilon := L - 1 \gt 0 \) to get \( |a_{n+1}/a_n| \gt L - \varepsilon = 1 \) eventually; if \( L = +\infty \), the ratios exceed \( 1 \) eventually by definition of divergence to \( +\infty \). Then \( (|a_n|)_{n \geq N} \) is strictly increasing, so bounded below by \( |a_N| \), which is positive since \( a_N \neq 0 \). A sequence bounded away from \( 0 \) cannot converge to \( 0 \). B

Since \( a_n \not\to 0 \), the series \( \sum_n a_n \) diverges, proving (ii). For (iii): \( \sum 1/n \) diverges and \( \sum 1/n^2 \) converges, yet both have \( L = \lim_n \frac{n}{n+1} = \lim_n \frac{n^2}{(n+1)^2} = 1 \).

The \( n \)-th term test (vanishing criterion): if \( \sum a_n \) converges then \( a_n = S_n - S_{n-1} \to S - S = 0 \); contrapositive gives divergence. Divergence of the harmonic series and convergence of \( \sum 1/n^2 \) are the standard facts proved in this unit by the Cauchy condensation test (or integral comparison). A

Result

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \lt 1 \;\Longrightarrow\; \sum_{n} |a_n| \lt \infty; \qquad L \gt 1 \;\Longrightarrow\; \sum_n a_n \text{ diverges}. \]

Reading. If, in the long run, each term is at most a fixed fraction of the previous one, the tail of the series is trapped underneath a convergent geometric series and the whole series converges absolutely. If instead the terms eventually grow, they cannot tend to zero, so the series has no chance of converging. Exactly at ratio \( 1 \), geometry says nothing and the test is silent.

Scope. Applies to any real or complex series whose terms are eventually nonzero and whose consecutive term ratios have a limit (finite or \( +\infty \)); the refined form replaces the limit by \( \limsup \lt 1 \) for convergence and \( \liminf \gt 1 \) for divergence. It decides nothing when the limit is \( 1 \), and in particular it can never establish conditional (non-absolute) convergence, nor settle any \( p \)-series.

Corollaries & converses

Power series. If \( c_n \neq 0 \) eventually and \( |c_{n+1}/c_n| \to \ell \), then \( \sum c_n z^n \) converges absolutely for \( |z| \lt 1/\ell \) and diverges for \( |z| \gt 1/\ell \) (with \( 1/0 := \infty \), \( 1/\infty := 0 \)): apply the theorem to \( a_n = c_n z^n \), whose ratio limit is \( \ell |z| \). This computes the radius of convergence whenever the coefficient ratios converge.
The exponential series. \( \sum_{n \geq 0} z^n / n! \) converges absolutely for every \( z \in \mathbb{C} \) (ratio \( |z|/(n+1) \to 0 \)); this is the standard construction of \( \exp \), and similarly of \( \sin \) and \( \cos \).
Growth comparison. If \( |a_{n+1}/a_n| \to L \lt 1 \) then \( a_n \to 0 \) — indeed \( a_n = \mathcal{O}(r^n) \) for any \( r \in (L, 1) \), by step 3. Useful facts like \( n^k x^n \to 0 \) for \( |x| \lt 1 \) and \( x^n / n! \to 0 \) follow instantly.
Converse fails. Absolute convergence does not force \( L \lt 1 \): \( \sum 1/n^2 \) converges absolutely with \( L = 1 \), and \( a_n = 2^{-n+(-1)^n} \) converges absolutely while \( L \) does not even exist. The ratio test is sufficient, never necessary.
Relation to the root test. Always \( \liminf_n \left| \frac{a_{n+1}}{a_n} \right| \leq \liminf_n |a_n|^{1/n} \leq \limsup_n |a_n|^{1/n} \leq \limsup_n \left| \frac{a_{n+1}}{a_n} \right| \), so whenever the ratio test decides, the root test decides identically — the root test is strictly stronger (see Problem 5).

Fails without

Drop \( L \lt 1 \), keep every individual ratio below \( 1 \): the harmonic series \( \sum 1/n \) has \( \left| \frac{a_{n+1}}{a_n} \right| = \frac{n}{n+1} \lt 1 \) for every \( n \), yet diverges. Pointwise ratio bounds below \( 1 \) are worthless; the hypothesis is that the limit stays a fixed distance below \( 1 \).
Allow \( L = 1 \): \( \sum 1/n^2 \) converges and \( \sum 1/n \) diverges, both with \( L = 1 \). Any claimed conclusion at \( L = 1 \) is refuted by one of the two.
Drop existence of the limit and use \( \limsup \gt 1 \) as a divergence criterion: for \( a_n = 2^{-n+(-1)^n} \) the ratios alternate between \( \tfrac18 \) and \( 2 \), so \( \limsup_n |a_{n+1}/a_n| = 2 \gt 1 \), yet \( \sum a_n \) converges absolutely since \( |a_n| \leq 2 \cdot 2^{-n} \). Divergence needs \( \liminf \gt 1 \) (or the ratios eventually \( \geq 1 \)), not \( \limsup \gt 1 \).
Drop absolute values: for real series with sign changes, the raw ratio \( a_{n+1}/a_n \) may converge to a number in \( (-1, 0) \) while telling you nothing directly; worse, students apply the test to \( a_{n+1}/a_n \to -2 \) and conclude divergence "since \( |{-2}| \gt 1 \)" — here that happens to be right, but only because the theorem is stated with \( \left| a_{n+1}/a_n \right| \); the test as proved concerns moduli, and conclusions must be routed through the modulus form.

Common errors

The \( L = 1 \) conclusion error: concluding convergence (or divergence) when the computed limit is \( 1 \). The test is genuinely silent there; switch to comparison, condensation, or Raabe's test.
The pointwise-ratio error: arguing "each ratio is less than \( 1 \), so the series converges". Refuted by \( \sum 1/n \). The limit must be \( \lt 1 \), or at least \( \limsup \lt 1 \) — equivalently the ratios must be \( \leq r \lt 1 \) uniformly from some point on.
Inverting the fraction: computing \( |a_n / a_{n+1}| \) and comparing to \( 1 \) the wrong way round — common with factorials, where the algebra invites cancellation in either order. Sanity-check: for the convergent \( \sum 2^{-n} \) the correct ratio is \( \tfrac12 \), not \( 2 \).
Misusing \( \limsup \gt 1 \): declaring divergence because the \( \limsup \) of the ratios exceeds \( 1 \). Divergence requires \( \liminf \gt 1 \) (or ratios eventually \( \geq 1 \)); see the third counterexample above.
Concluding only "convergence": the test delivers absolute convergence, which is strictly stronger (it survives rearrangement, by the rearrangement theorem for absolutely convergent series). Conversely, the test can never certify a conditionally convergent series such as \( \sum (-1)^n / n \): there \( L = 1 \).
Forgetting the terms may vanish: applying the test to a series with infinitely many zero terms (e.g. a power series with gaps, \( \sum z^{n^2} \)); the ratios are undefined and the root test or direct comparison must be used instead.

Discussion

The test is traditionally credited to Jean le Rond d'Alembert (1768), who used ratio comparisons to discuss the binomial series, and it is often called d'Alembert's ratio test. Cauchy's Cours d'analyse (1821) placed it, together with his root test, inside the first systematic theory of convergence, and the modern \( \limsup \) formulations are essentially Cauchy's. The proof above is a faithful record of what the test really is: a machine for manufacturing a comparison with a geometric series. Nothing about series is used beyond the comparison principle and the convergence of \( \sum r^n \) for \( 0 \leq r \lt 1 \).

That observation explains both the power and the limits of the test. It is decisive exactly for series of geometric type — terms built from powers, factorials, and products, where consecutive ratios simplify dramatically. It is helpless for series of polynomial type such as \( \sum n^{-p} \), because \( (n/(n+1))^p \to 1 \) for every \( p \): the geometric yardstick is too coarse to distinguish polynomial decay rates. The finer tests that resolve the \( L = 1 \) boundary — Raabe's, Bertrand's, Gauss's — all work by measuring how fast the ratio approaches \( 1 \).

The comparison with the root test is worth internalising. The chain \( \liminf |a_{n+1}/a_n| \leq \liminf |a_n|^{1/n} \leq \limsup |a_n|^{1/n} \leq \limsup |a_{n+1}/a_n| \) (a standard lemma, proved by the same geometric-domination argument as step 3) shows the root test succeeds whenever the ratio test does, and Problem 5 shows the converse fails. In the theory of power series this is why the Cauchy–Hadamard formula for the radius of convergence uses \( \limsup |c_n|^{1/n} \), which always exists in \( [0,\infty] \), rather than a ratio: the ratio version is a convenience valid only when the coefficient ratios happen to converge.

All of these tests are shadows of a single result, Kummer's test (1835, simplified by Dini): for a series of positive terms \( a_n \), if there exist positive numbers \( p_n \) and a constant \( c \gt 0 \) with \( p_n \frac{a_n}{a_{n+1}} - p_{n+1} \geq c \) eventually, then \( \sum a_n \) converges; and if \( p_n \frac{a_n}{a_{n+1}} - p_{n+1} \leq 0 \) eventually with \( \sum 1/p_n \) divergent, then \( \sum a_n \) diverges. Taking \( p_n = 1 \) recovers the ratio test; \( p_n = n \) gives Raabe's test, sensitive to ratios of the form \( 1 - \frac{\alpha}{n} + \mathcal{O}(n^{-2}) \) (convergence iff \( \alpha \gt 1 \), matching the \( p \)-series threshold); \( p_n = n \log n \) gives Bertrand's test, one level deeper. Remarkably, Kummer's test is in principle complete: for every convergent positive series a valid choice of \( (p_n) \) exists — but the choice requires knowing the tail sums, so the hierarchy of practical tests is genuinely infinite, and no single effective test decides all series. Common misconceptions. The ratio test is not "the definition of convergence for series with factorials", and \( L \lt 1 \) is not necessary for convergence — it is a sufficient condition only. Nor does \( L \lt 1 \) say the terms are eventually decreasing by the factor \( L \); they are eventually decreasing by any factor \( r \) strictly between \( L \) and \( 1 \), which is weaker and is exactly what the proof uses.

Worked examples

Example 1. Show that the exponential series \( \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} \) converges absolutely for every \( x \in \mathbb{R} \).

If \( x = 0 \) the series is \( 1 + 0 + 0 + \cdots \), trivially absolutely convergent. Assume \( x \neq 0 \) and set \( a_n = \dfrac{x^n}{n!} \); then \( a_n \neq 0 \) for all \( n \), so the ratio-test hypotheses can be checked.

Disposing of the degenerate case; nonvanishing of the terms since \( x \neq 0 \) and \( n! \gt 0 \). A

\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n} \right| = \frac{|x|}{n+1}. \]

Algebra: \( (n+1)! = (n+1) \cdot n! \) and \( x^{n+1}/x^n = x \). A

\[ L = \lim_{n \to \infty} \frac{|x|}{n+1} = 0 \lt 1. \]

The numerator is a fixed constant and \( n + 1 \to \infty \); Archimedean property of \( \mathbb{R} \). Note \( L = 0 \) regardless of how large \( |x| \) is — the factorial always wins. B

\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} \text{ converges absolutely for every } x \in \mathbb{R} \text{ (and every } x \in \mathbb{C}). \]

Reading. The ratio test with \( L = 0 \): factorial growth in the denominator crushes any fixed geometric growth in the numerator, so \( \exp \) is defined on the whole line (the same computation over \( \mathbb{C} \) gives the whole plane).

Scope. This is the standard construction of the exponential function in this unit; the identical argument handles \( \sin \) and \( \cos \).

Example 2. Determine whether \( \displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n} \) converges.

Set \( a_n = \dfrac{n!}{n^n} \gt 0 \). Then \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{(n+1) \, n^n}{(n+1)^{n+1}} = \left( \frac{n}{n+1} \right)^{n}. \]

Algebra: cancel \( n! \) into \( (n+1)! \), then cancel one factor of \( n + 1 \). All terms are positive so moduli are redundant. A

\[ \left( \frac{n}{n+1} \right)^{n} = \frac{1}{\left(1 + \frac{1}{n}\right)^{n}} \;\longrightarrow\; \frac{1}{e} \quad (n \to \infty). \]

The standard limit \( \lim_{n} \left(1 + \frac{1}{n}\right)^n = e \), proved earlier in this unit from monotonicity and boundedness of the sequence, together with continuity of \( t \mapsto 1/t \) at \( e \neq 0 \) (algebra of limits). B

\[ L = \frac{1}{e} \approx 0.3679 \lt 1. \]

Since \( e \gt 1 \) (indeed \( e \gt 2 \), from the first two terms of its defining series), \( 1/e \lt 1 \). Ratio test applies with \( L \lt 1 \). A

\[ \sum_{n=1}^{\infty} \frac{n!}{n^n} \text{ converges (absolutely).} \]

Reading. \( n^n \) outgrows \( n! \) at a per-step rate approaching \( e \), so the terms decay essentially like \( e^{-n} \) — geometric decay, which the ratio test detects immediately.

Scope. A comparison-test proof is possible (\( n!/n^n \leq 2/n^2 \) with effort), but the ratio computation is a two-line cancellation; this is the test operating in its natural habitat of factorials and powers.

Problems

Decide whether \( \displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} \) converges.
Solution

Set \( a_n = n/2^n \gt 0 \). Then \[ \frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{1}{2} \cdot \frac{n+1}{n} = \frac{1}{2}\left(1 + \frac{1}{n}\right) \;\longrightarrow\; \frac{1}{2}. \] Since \( L = \tfrac12 \lt 1 \), the series converges absolutely by the ratio test. (Its value, not required here, is \( 2 \).)
Decide whether \( \displaystyle\sum_{n=0}^{\infty} \frac{(n!)^2}{(2n)!} \) converges.
Solution

Set \( a_n = \dfrac{(n!)^2}{(2n)!} \gt 0 \). Then \[ \frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2} = \frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{(n+1)^2}{2(n+1)(2n+1)} = \frac{n+1}{2(2n+1)}. \] As \( n \to \infty \) this tends to \( \dfrac{1}{4} \lt 1 \), so the series converges absolutely by the ratio test. (Note \( a_n = 1/\binom{2n}{n} \); the test shows the central binomial coefficients grow essentially like \( 4^n \).)
Find all \( x \in \mathbb{R} \) for which \( \displaystyle\sum_{n=1}^{\infty} n x^n \) converges.
Solution

For \( x = 0 \) every term vanishes and the series converges trivially. For \( x \neq 0 \), set \( a_n = n x^n \neq 0 \); then \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)|x|^{n+1}}{n |x|^n} = |x| \left( 1 + \frac{1}{n} \right) \;\longrightarrow\; |x|. \] By the ratio test the series converges absolutely when \( |x| \lt 1 \) and diverges when \( |x| \gt 1 \). At the boundary the test is silent, so check directly: if \( |x| = 1 \) then \( |a_n| = n \not\to 0 \), and the series diverges by the \( n \)-th term test. Hence the series converges exactly for \( x \in (-1, 1) \), and the convergence there is absolute.
Let \( p \gt 0 \). Show that the ratio test gives no information about \( \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p} \) for any value of \( p \), and explain what this says about the strength of the test.
Solution

With \( a_n = n^{-p} \gt 0 \), \[ \frac{a_{n+1}}{a_n} = \left( \frac{n}{n+1} \right)^p = \left( 1 + \frac{1}{n} \right)^{-p} \;\longrightarrow\; 1^{-p} = 1, \] using continuity of \( t \mapsto t^{-p} \) at \( t = 1 \). So \( L = 1 \) for every \( p \gt 0 \), and the ratio test is inconclusive in every case — even though the true behaviour depends on \( p \): the series converges for \( p \gt 1 \) and diverges for \( p \leq 1 \) (Cauchy condensation: \( \sum 2^k \cdot 2^{-kp} = \sum (2^{1-p})^k \) is geometric, converging iff \( 2^{1-p} \lt 1 \) iff \( p \gt 1 \)). Conclusion: the ratio test measures only geometric-scale decay. All polynomial decay rates are invisible to it — they compress into the single boundary point \( L = 1 \). Distinguishing them requires finer tests (comparison, condensation, Raabe).
Let \( a_n = 2^{-n + (-1)^n} \) for \( n \geq 1 \). (a) Show that \( \lim_n |a_{n+1}/a_n| \) does not exist and that \( \limsup_n |a_{n+1}/a_n| = 2 \), \( \liminf_n |a_{n+1}/a_n| = \tfrac18 \), so that no form of the ratio test applies. (b) Prove nonetheless that \( \sum a_n \) converges absolutely. (c) Show that the root test succeeds, and conclude that the root test is strictly stronger than the ratio test.
Solution

(a) All \( a_n \gt 0 \). Compute \[ \frac{a_{n+1}}{a_n} = 2^{\,[-(n+1) + (-1)^{n+1}] - [-n + (-1)^n]} = 2^{-1 + (-1)^{n+1} - (-1)^n}. \] If \( n \) is even, \( (-1)^{n+1} - (-1)^n = -1 - 1 = -2 \), giving ratio \( 2^{-3} = \tfrac18 \); if \( n \) is odd, \( (-1)^{n+1} - (-1)^n = 1 - (-1) = 2 \), giving ratio \( 2^{1} = 2 \). The ratios take the values \( \tfrac18 \) and \( 2 \) alternately, so the limit does not exist; the \( \limsup \) is \( 2 \gt 1 \) and the \( \liminf \) is \( \tfrac18 \lt 1 \). The limit form fails (no limit), the refined form fails (\( \limsup \not\lt 1 \) and \( \liminf \not\gt 1 \)): the ratio test, in every version, is silent.

(b) Since \( (-1)^n \leq 1 \), we have \( 0 \lt a_n = 2^{-n+(-1)^n} \leq 2^{-n+1} = 2 \cdot 2^{-n} \). The series \( \sum 2 \cdot 2^{-n} \) is a convergent geometric series (ratio \( \tfrac12 \lt 1 \)), so by the comparison test \( \sum a_n \) converges; as the terms are positive, the convergence is absolute.

(c) \[ a_n^{1/n} = 2^{\frac{-n + (-1)^n}{n}} = 2^{-1 + \frac{(-1)^n}{n}} \;\longrightarrow\; 2^{-1} = \frac{1}{2}, \] since \( (-1)^n / n \to 0 \) and \( t \mapsto 2^t \) is continuous. Thus \( \limsup_n a_n^{1/n} = \tfrac12 \lt 1 \) and the root test proves absolute convergence directly. Combined with the general inequality \( \limsup_n |a_n|^{1/n} \leq \limsup_n |a_{n+1}/a_n| \) (which shows the root test decides whenever the ratio test does), this example shows the root test decides strictly more series: the root test is strictly stronger.