The Root Test

Suppose you are handed a series and asked whether it converges. Its terms involve nested powers, factorials disguised as exponentials, or coefficients that oscillate wildly from term to term. The Ratio Test often chokes on such examples. The ratio of consecutive terms fails to settle down to a single limit. The Root Test resolves exactly this difficulty. Instead of comparing consecutive terms, it examines the long-run growth rate of the terms themselves. It does this by taking nn-th roots. It uses the Limit Superior and Limit Inferior rather than an ordinary limit. This ensures no series with erratic behavior escapes its grasp. The idea traces back to Cauchy. He realized that a series ∑an\sum a_n behaves, in the limit, like a geometric series. Its ratio is r=lim sup⁡n→∞∣an∣1/nr = \limsup_{n\to\infty} |a_n|^{1/n}. If r<1r<1, the terms eventually decay at least geometrically fast. This guarantees convergence. If r>1r>1, they eventually grow without bound. This forces divergence. The genius of using lim sup⁡\limsup instead of an ordinary limit is that it never fails to exist. It is always a well-defined extended real number for any bounded or unbounded sequence of nonnegative reals. So the test applies to every series, even ones whose term-by-term behavior is chaotic. This makes the Root Test strictly more powerful than the Ratio Test in a precise sense. Whenever the ratio test's limit exists, the root test's limsup exists and equals it. But there are series where the root test succeeds and the ratio test gives no information at all. This extra strength is not a curiosity. It becomes indispensable when we study Power Series and Radius of Convergence. There, the Cauchy–Hadamard formula for the radius of convergence is literally the Root Test in disguise. It is applied uniformly across an entire family of series indexed by xx. This article gives the precise statement. It offers a complete proof by direct comparison with geometric series. It gives a worked demonstration of the test's superiority over the ratio test. It also carefully discusses the one case — r=1r=1 — where the test is silent. No amount of cleverness can fix that.

Key takeaways
  • The Root Test studies ρ=lim sup⁡n→∞∣an∣1/n\rho = \limsup_{n\to\infty} |a_n|^{1/n} for a series ∑an\sum a_n.
  • If ρ<1\rho < 1 the series converges absolutely; if ρ>1\rho > 1 it diverges; if ρ=1\rho = 1 the test is inconclusive.
  • Uses lim sup⁡\limsup, not an ordinary limit, so it applies to every real or complex sequence, oscillating or not.
  • Strictly stronger than the Ratio Test: whenever the ratio test's limit exists, the root test agrees with it.
  • The proof is a direct comparison against a geometric series, using the defining property of lim sup⁡\limsup.
  • Key idea: the terms of a convergent series must eventually shrink at least as fast as a geometric sequence, and lim sup⁡∣an∣1/n\limsup |a_n|^{1/n} is exactly the threshold rate that detects this.

Where this lives today

The Root Test's most direct descendant is the Cauchy–Hadamard formula for the radius of convergence of a power series ∑cnxn\sum c_n x^n: the radius is R=1/lim sup⁡n∣cn∣1/nR = 1/\limsup_n |c_n|^{1/n}, obtained by applying the Root Test to the series with terms an=cnxna_n = c_n x^n and solving the inequality ρ<1\rho<1 for xx — this is the direct bridge to Power Series and Radius of Convergence. In complex analysis, the same idea underlies convergence criteria for Laurent series and the classification of singularities. In numerical analysis and dynamical systems, root-type growth-rate tests appear when bounding the growth of iterated sequences or verifying stability of numerical schemes, where lim sup⁡\limsup of nn-th roots of error terms measures asymptotic contraction rates. In information theory and combinatorics, exponential generating function techniques routinely invoke root-test-style asymptotics (often refined via singularity analysis) to extract growth rates of combinatorial sequences from their generating functions. In probability theory, the root test's logic reappears in large deviations theory, where lim sup⁡\limsup of normalized log-probabilities (a close cousin of ∣an∣1/n|a_n|^{1/n}) governs exponential decay rates of rare-event probabilities.

Ask the tutor: Can you show me how the Root Test's proof changes if we allow ana_n to be vectors in a normed space instead of real or complex numbers?

DefinitionThe Root Test

Let ∑n=1∞an\sum_{n=1}^{\infty} a_n be a series of real or complex numbers. Define

ρ=lim sup⁡n→∞∣an∣1/n∈[0,∞].\rho = \limsup_{n\to\infty} |a_n|^{1/n} \in [0,\infty].

The Root Test asserts: 1. If ρ<1\rho < 1, the series ∑an\sum a_n converges absolutely. 2. If ρ>1\rho > 1 (including ρ=∞\rho = \infty), the series ∑an\sum a_n diverges. 3. If ρ=1\rho = 1, the test is inconclusive: series with ρ=1\rho=1 exist that converge, and series with ρ=1\rho=1 exist that diverge.

Intuition

Think of ∣an∣1/n|a_n|^{1/n} as measuring the "average per-step growth factor" of the term ana_n, the way a bank account's nn-year growth is summarized by an annual rate. If eventually this rate is bounded above by some c<1c<1, then ∣an∣|a_n| is eventually bounded by cnc^n, and a geometric series with ratio c<1c<1 converges — so by the Comparison Test, so does ∑an\sum a_n. The role of lim sup⁡\limsup is to extract exactly this "eventual ceiling" on the growth rate, ignoring finitely many wild terms and even ignoring infinitely many terms that behave better than the ceiling, as long as they never permanently exceed it. Conversely, if ∣an∣1/n|a_n|^{1/n} exceeds 11 infinitely often (which is what ρ>1\rho > 1 means for the limsup), then ∣an∣>1|a_n| > 1 for infinitely many nn, so the terms cannot tend to 00, and the series diverges by the divergence test. The boundary case ρ=1\rho = 1 is genuinely ambiguous: it is the threshold between geometric-like decay and non-decay, and both convergent and divergent series can sit exactly on this threshold, as the classical examples ∑1/n2\sum 1/n^2 (converges) and ∑1/n\sum 1/n (diverges) show — both have ρ=1\rho=1.

Figure
The threshold ρ=1\rho=1 separates geometric-like decay from growth
n |aₙ|^(1/n) ρ = 1 ρ < 1: converges ρ = 1: inconclusive ρ > 1: diverges
TheoremRoot Test

Let ∑n=1∞an\sum_{n=1}^{\infty} a_n be a series of complex numbers, and set

ρ=lim sup⁡n→∞∣an∣1/n.\rho = \limsup_{n\to\infty} |a_n|^{1/n}.

Then: (a) If ρ<1\rho < 1, then ∑n=1∞an\sum_{n=1}^\infty a_n converges absolutely. (b) If ρ>1\rho > 1 (or ρ=∞\rho = \infty), then ∑n=1∞an\sum_{n=1}^\infty a_n diverges, in fact an↛0a_n \not\to 0. (c) If ρ=1\rho = 1, the test gives no information: there exist convergent and divergent series with this value of ρ\rho.

Proof

Technique: direct comparison with a geometric series, using the epsilon-characterization of lim sup⁡\limsup. We first record the defining property of limit superior that we will use (this is part of the standard theory of Limit Superior and Limit Inferior, assumed known): for a sequence of nonnegative reals (xn)(x_n) with lim sup⁡nxn=L∈[0,∞]\limsup_n x_n = L \in [0,\infty],

  • (P1) for every ε>0\varepsilon > 0, there exists NN such that xn<L+εx_n < L + \varepsilon for all n≥Nn \geq N (holds when L<∞L < \infty);
  • (P2) for every ε>0\varepsilon > 0 and every NN, there exists n≥Nn \geq N with xn>L−εx_n > L - \varepsilon (holds when L>0L > 0, in particular whenever L>0L > 0 this gives infinitely many nn with xnx_n close to or above L−εL-\varepsilon).

Part (a): ρ<1  ⟹  \rho < 1 \implies absolute convergence. Choose rr with ρ<r<1\rho < r < 1; such rr exists since ρ<1\rho < 1 (e.g. r=ρ+12r = \frac{\rho+1}{2}). Let ε=r−ρ>0\varepsilon = r - \rho > 0. By (P1) applied to xn=∣an∣1/nx_n = |a_n|^{1/n} and L=ρL = \rho, there exists N∈NN \in \mathbb{N} such that

∣an∣1/n<ρ+ε=rfor all n≥N.|a_n|^{1/n} < \rho + \varepsilon = r \quad \text{for all } n \geq N.

Raising both sides to the nn-th power (valid since both sides are nonnegative and t↦tnt \mapsto t^n is increasing on [0,∞)[0,\infty)):

∣an∣<rnfor all n≥N.|a_n| < r^n \quad \text{for all } n \geq N.

Since 0≤r<10 \le r < 1, the geometric series ∑n=N∞rn\sum_{n=N}^{\infty} r^n converges (its partial sums are rN−rM+11−r→rN1−r\frac{r^N - r^{M+1}}{1-r} \to \frac{r^N}{1-r} as M→∞M\to\infty). By the Comparison Test applied to the nonnegative series ∑∣an∣\sum |a_n| and ∑rn\sum r^n (term-by-term comparison for n≥Nn \ge N, with the finitely many terms n<Nn<N not affecting convergence), the series ∑n=N∞∣an∣\sum_{n=N}^\infty |a_n| converges. Adding back the finitely many terms n=1,…,N−1n = 1,\dots,N-1 (a finite sum, hence finite) shows ∑n=1∞∣an∣\sum_{n=1}^{\infty} |a_n| converges. That is, ∑an\sum a_n converges absolutely.

Part (b): ρ>1  ⟹  \rho > 1 \implies divergence. Case ρ>1\rho > 1 finite: apply (P2) with L=ρ>1L = \rho > 1 and ε=ρ−1>0\varepsilon = \rho - 1 > 0. For every NN, there exists n≥Nn \geq N with

∣an∣1/n>ρ−ε=1,|a_n|^{1/n} > \rho - \varepsilon = 1,

hence ∣an∣>1n=1|a_n| > 1^n = 1 for that nn. Since this holds for infinitely many nn (we may repeat the argument with NN replaced by any larger value, producing infinitely many such indices), the sequence (an)(a_n) has a subsequence bounded below by 11 in absolute value, so an↛0a_n \not\to 0. Case ρ=∞\rho = \infty: for every M>0M > 0 and every NN, there exists n≥Nn \ge N with ∣an∣1/n>M|a_n|^{1/n} > M (this is the defining property of lim sup⁡=∞\limsup = \infty: the sequence is not eventually bounded above by any MM). Taking M=1M=1 gives the same conclusion as above: infinitely many nn with ∣an∣>1|a_n| > 1, so an↛0a_n \not\to 0. In either case, a necessary condition for convergence of ∑an\sum a_n fails: if ∑an\sum a_n converged, its terms would satisfy an→0a_n \to 0 (since partial sums Sn→SS_n \to S implies an=Sn−Sn−1→S−S=0a_n = S_n - S_{n-1} \to S - S = 0). Since an↛0a_n \not\to 0, the series diverges.

Part (c): ρ=1\rho = 1 is inconclusive. We exhibit two series both with ρ=1\rho = 1, one convergent and one divergent. Divergent example: an=1/na_n = 1/n. Then ∣an∣1/n=n−1/n=e−ln⁡nn|a_n|^{1/n} = n^{-1/n} = e^{-\frac{\ln n}{n}}. Since ln⁡nn→0\frac{\ln n}{n} \to 0 as n→∞n \to \infty (standard limit, e.g. by L'Hôpital or by comparing ln⁡n\ln n to n\sqrt n), we get ∣an∣1/n→e0=1|a_n|^{1/n} \to e^0 = 1, so lim sup⁡=lim⁡=1\limsup = \lim = 1. But ∑1/n\sum 1/n is the harmonic series, which diverges (standard fact, e.g. via the Comparison Test with the integral ∫1∞dx/x\int_1^\infty dx/x, or Cauchy condensation). Convergent example: an=1/n2a_n = 1/n^2. Then ∣an∣1/n=n−2/n=(n−1/n)2→12=1|a_n|^{1/n} = n^{-2/n} = \left(n^{-1/n}\right)^2 \to 1^2 = 1 by the same limit as above and continuity of squaring. So again ρ=1\rho = 1. But ∑1/n2\sum 1/n^2 converges (standard fact, e.g. comparison with the telescoping series ∑(1n−1−1n)\sum \left(\frac1{n-1}-\frac1n\right) for n≥2n\ge 2, or the integral test). Both series have ρ=1\rho = 1, yet one converges and the other diverges. Hence no conclusion about convergence or divergence can be drawn purely from the value ρ=1\rho = 1; the test is uninformative in this case. ■\blacksquare

∎

Worked examples

Example 1: a series where the Ratio Test is silent but the Root Test succeeds. Consider the series ∑n=1∞an\sum_{n=1}^\infty a_n with terms defined by interleaving two geometric-like patterns:

an={12nn odd13n−1n evena_n = \begin{cases} \dfrac{1}{2^n} & n \text{ odd} \\[4pt] \dfrac{1}{3^{n-1}} & n \text{ even} \end{cases}

So explicitly a1=1/2, a2=1/3, a3=1/8, a4=1/27, a5=1/32, a6=1/243,…a_1 = 1/2,\ a_2 = 1/3,\ a_3 = 1/8,\ a_4 = 1/27,\ a_5 = 1/32,\ a_6 = 1/243,\dots Ratio test attempt. Compute an+1/ana_{n+1}/a_n for consecutive terms. From odd n=2k−1n=2k-1 to even n=2kn=2k:

a2ka2k−1=1/32k−11/22k−1=(23)2k−1→0.\frac{a_{2k}}{a_{2k-1}} = \frac{1/3^{2k-1}}{1/2^{2k-1}} = \left(\frac{2}{3}\right)^{2k-1} \to 0.

From even n=2kn=2k to odd n=2k+1n=2k+1:

a2k+1a2k=1/22k+11/32k−1=32k−122k+1=16(32)2k→∞.\frac{a_{2k+1}}{a_{2k}} = \frac{1/2^{2k+1}}{1/3^{2k-1}} = \frac{3^{2k-1}}{2^{2k+1}} = \frac{1}{6}\left(\frac{3}{2}\right)^{2k} \to \infty.

The ratio an+1/ana_{n+1}/a_n oscillates between values tending to 00 and values tending to ∞\infty; it has no limit, so the Ratio Test simply does not apply (there is no single limiting ratio to test). Root test. Compute ∣an∣1/n|a_n|^{1/n} along each subsequence. For odd nn: an=2−na_n = 2^{-n}, so ∣an∣1/n=2−1=1/2|a_n|^{1/n} = 2^{-1} = 1/2. For even nn: an=3−(n−1)a_n = 3^{-(n-1)}, so ∣an∣1/n=3−(n−1)/n=3−1+1/n→3−1=1/3|a_n|^{1/n} = 3^{-(n-1)/n} = 3^{-1 + 1/n} \to 3^{-1} = 1/3 as n→∞n\to\infty. So the sequence ∣an∣1/n|a_n|^{1/n} has two subsequential limits, 1/21/2 (from odd nn, exactly, for every odd nn) and a sequence converging to 1/31/3 (from even nn). Hence

lim sup⁡n→∞∣an∣1/n=max⁡(12, 13)=12.\limsup_{n\to\infty} |a_n|^{1/n} = \max\left(\tfrac12,\ \tfrac13\right) = \tfrac12.

Since ρ=1/2<1\rho = 1/2 < 1, the Root Test guarantees ∑an\sum a_n converges absolutely — a conclusion the Ratio Test could not reach, since it never even produced a candidate limit.

Example 2: a series requiring careful handling of a non-power prefactor. Consider

∑n=1∞an,an=(n+12n+5)n⋅n7.\sum_{n=1}^{\infty} a_n, \qquad a_n = \left(\frac{n+1}{2n+5}\right)^n \cdot n^{7}.

The polynomial factor n7n^7 has no bearing on exponential-scale growth, but let us verify this rigorously rather than wave it away. We compute ∣an∣1/n=(n+12n+5)⋅(n7)1/n=n+12n+5⋅n7/n|a_n|^{1/n} = \left(\dfrac{n+1}{2n+5}\right) \cdot (n^{7})^{1/n} = \dfrac{n+1}{2n+5}\cdot n^{7/n}. First, lim⁡n→∞n+12n+5=lim⁡n→∞1+1/n2+5/n=12\displaystyle\lim_{n\to\infty}\frac{n+1}{2n+5} = \lim_{n\to\infty}\frac{1+1/n}{2+5/n} = \frac{1}{2}, by dividing numerator and denominator by nn and using that 1/n→01/n \to 0. Second, n7/n=e7ln⁡nn→e0=1n^{7/n} = e^{\frac{7\ln n}{n}} \to e^{0} = 1, since ln⁡nn→0\frac{\ln n}{n}\to 0 as used above. By the product rule for limits (both factors converge, so the product converges to the product of the limits — this uses that ordinary limits, when they exist, coincide with lim sup⁡\limsup, a basic fact about Limit Superior and Limit Inferior):

ρ=lim⁡n→∞∣an∣1/n=12⋅1=12.\rho = \lim_{n\to\infty} |a_n|^{1/n} = \frac12 \cdot 1 = \frac12.

Since ρ=1/2<1\rho = 1/2 < 1, the Root Test gives absolute convergence of ∑an\sum a_n. Note the polynomial factor n7n^7 genuinely does not affect ρ\rho because n7/n→1n^{7/n}\to 1 — this is the general phenomenon that polynomial (or even sub-exponential) prefactors never change the value of lim sup⁡∣an∣1/n\limsup|a_n|^{1/n}, only true exponential-scale factors do.

CounterexampleWhy we need lim sup⁡\limsup, not an ordinary limit — and why ρ=1\rho=1 truly cannot be resolved in general

One might be tempted to state the Root Test using lim⁡n→∞∣an∣1/n\lim_{n\to\infty}|a_n|^{1/n} instead of lim sup⁡\limsup, on the grounds that "in practice the limit usually exists." This weakened hypothesis breaks the test's universal applicability: Example 1 above is precisely a series for which lim⁡n→∞∣an∣1/n\lim_{n\to\infty}|a_n|^{1/n} does not exist (the sequence ∣an∣1/n|a_n|^{1/n} has two distinct subsequential limits, 1/21/2 and 1/31/3, so no single limit exists), yet the series still converges and lim sup⁡\limsup still correctly detects this. A test phrased with an ordinary limit would simply not apply to this series at all, giving no verdict, even though the correct verdict (convergence) is perfectly accessible via lim sup⁡\limsup. This also illustrates why the boundary value ρ=1\rho = 1 cannot be strengthened into a real conclusion by any modification of the test: the two examples ∑1/n\sum 1/n and ∑1/n2\sum 1/n^2 genuinely share the same value ρ=1\rho=1 while having opposite convergence behavior. No test depending only on the single real number ρ\rho can distinguish these two series, since they produce the identical output ρ=1\rho=1. Any correct decision in the boundary case necessarily requires extra information beyond ρ\rho — for instance, the sharper asymptotic rate at which ∣an∣1/n|a_n|^{1/n} approaches 11, which is exactly the content of finer tools like Raabe's test.

Scope, edge cases & common traps

SituationWhat happensWhy
ρ<1\rho < 1Converges absolutelyEventually dominated by geometric series with ratio r<1r<1
ρ>1\rho > 1, including ρ=∞\rho = \inftyDivergesTerms don't tend to 00
ρ=1\rho = 1InconclusiveBoth ∑1/n\sum 1/n (diverges) and ∑1/n2\sum 1/n^2 (converges) achieve ρ=1\rho=1
lim⁡∣an∣1/n\lim |a_n|^{1/n} existsRoot test = Ratio test's verdict (when ratio test's limit also exists)Ordinary limits are special cases of limsup
lim⁡∣an∣1/n\lim |a_n|^{1/n} does not exist but lim sup⁡\limsup doesRoot test still workslimsup always exists in [0,∞][0,\infty] for any sequence bounded below by 00
Polynomial or sub-exponential prefactors (nkn^k, log⁡n\log n, etc.)Do not change ρ\rhonk/n→1n^{k/n}\to 1 for any fixed real kk

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Post-revolutionary French mathematics, École Polytechnique

Cauchy was among the founders of rigorous analysis, replacing the informal 18th-century treatment of limits and infinite series with precise epsilon-based definitions. Working during a period when calculus's foundations were being systematically rebuilt after the looseness of Euler and the Bernoullis, Cauchy introduced the modern definitions of limit, continuity, and convergence used to this day, and it is in this spirit that the Root Test (sometimes called the Cauchy Root Test or Cauchy's Criterion for series) originates. His 1821 textbook Cours d'Analyse, written for his students at the École Polytechnique, is widely regarded as the first truly rigorous calculus text and contains the root test essentially in its modern form, well before the notion of limsup was formalized with the notation we use today.

History
- 1821 — Cauchy publishes Cours d'Analyse, formulating the root test for series convergence using what we would now phrase as lim sup⁡\limsup.
- 1830s–1840s — Cauchy and others develop rigorous convergence tests (ratio, root, integral) as part of the broader "rigorization" movement in analysis.
- 1892 — Jacques Hadamard applies the Cauchy root criterion to power series, producing the Cauchy–Hadamard formula for the radius of convergence.
- Late 19th century — the modern lim inf⁡/lim sup⁡\liminf/\limsup notation and theory, building on Cauchy's and Bolzano–Weierstrass-era ideas, gives the test its fully general modern statement.
- 20th century — the root test becomes a standard fixture of undergraduate analysis courses (Rudin, Apostol, and successors), always paired with the ratio test as complementary tools.
Why does the Root Test use lim sup⁡\limsup instead of an ordinary limit?

Because lim sup⁡n→∞∣an∣1/n\limsup_{n\to\infty}|a_n|^{1/n} is guaranteed to exist as an extended real number in [0,∞][0,\infty] for absolutely any sequence of terms, while the ordinary limit lim⁡n∣an∣1/n\lim_n |a_n|^{1/n} may fail to exist even for perfectly convergent series (see the worked interleaved example). Using lim sup⁡\limsup means the test is never "inapplicable due to non-existence of a limit" — it always produces a definite number ρ\rho, and the only genuinely ambiguous case is the boundary value ρ=1\rho=1. This makes the Root Test logically complete in a way the naive limit-based version is not, and it is precisely why the rigorous statement always uses lim sup⁡\limsup.

How does the Root Test compare to the Ratio Test in practice?

Whenever lim⁡n∣an+1/an∣\lim_n |a_{n+1}/a_n| exists, it can be shown to equal lim⁡n∣an∣1/n\lim_n |a_n|^{1/n} (a classical fact sometimes proved via the Stolz–Cesàro theorem), so in that overlap case both tests give the identical verdict. The Root Test is strictly more general: it is defined via lim sup⁡\limsup and hence applies even when the ratio an+1/ana_{n+1}/a_n oscillates or is undefined (e.g. when infinitely many an=0a_n=0). In practice, the Ratio Test is often computationally easier for terms built from factorials or simple exponentials, while the Root Test is preferred for terms that are themselves nn-th powers, like an=cnna_n = c_n^n, since the root immediately simplifies.

Does the Root Test tell us about conditional convergence?

No. The Root Test only ever certifies absolute convergence (when ρ<1\rho<1) or divergence (when ρ>1\rho>1). It says nothing about series that converge conditionally but not absolutely, such as the alternating harmonic series ∑(−1)n/n\sum (-1)^n/n. For that series, ρ=lim sup⁡∣(−1)n/n∣1/n=lim⁡n−1/n=1\rho = \limsup |{(-1)^n/n}|^{1/n} = \lim n^{-1/n} = 1, so the Root Test is inconclusive — appropriately so, since the series converges only conditionally, a phenomenon entirely outside what a test for absolute convergence can be expected to detect.

Can ρ\rho ever be negative or can the test apply to complex series directly?

ρ\rho is always a limit of nonnegative real numbers ∣an∣1/n≥0|a_n|^{1/n}\ge 0, so ρ∈[0,∞]\rho\in[0,\infty] always; it is never negative. The test applies directly to complex series ∑an\sum a_n with an∈Ca_n\in\mathbb{C}, since ∣an∣|a_n| (the complex modulus) is well-defined and nonnegative regardless. The conclusion "ρ<1⇒\rho<1 \Rightarrow absolute convergence" is exactly as strong in the complex case, since absolute convergence of a complex series (meaning ∑∣an∣<∞\sum |a_n|<\infty) implies convergence of the complex series itself, by completeness of C\mathbb C.

What happens if infinitely many terms ana_n are exactly zero?

This causes no issue: if an=0a_n=0, define ∣an∣1/n=0|a_n|^{1/n}=0 by convention (consistent with 01/n=00^{1/n}=0 for n≥1n\ge 1), and the limsup is computed over the full sequence including these zero values. Zero terms only ever help convergence (they contribute nothing to the sum) and can only push lim sup⁡\limsup downward or leave it unchanged, so the test remains valid and meaningful even for series with infinitely many zero terms, such as sparse series arising from lacunary power series.

Is there a version of the Root Test with lim inf⁡\liminf instead of lim sup⁡\limsup?

Not for a standard convergence test, because lim inf⁡∣an∣1/n<1\liminf|a_n|^{1/n}<1 does not guarantee that all but finitely many terms are eventually small — it only guarantees that infinitely many terms are small, which is too weak to bound the tail of the series. The correct and only valid formulation genuinely requires lim sup⁡\limsup, since convergence requires control of every sufficiently large term, and lim sup⁡\limsup is exactly the tool that captures "eventually always below a bound," while lim inf⁡\liminf only captures "infinitely often below a bound."

How does the Root Test connect to power series and radius of convergence?

Given a power series ∑cnxn\sum c_n x^n, applying the Root Test to the terms an=cnxna_n = c_n x^n gives ρ(x)=∣x∣⋅lim sup⁡n∣cn∣1/n\rho(x) = |x|\cdot \limsup_n|c_n|^{1/n}. Setting L=lim sup⁡n∣cn∣1/nL=\limsup_n |c_n|^{1/n}, the series converges absolutely when ∣x∣<1/L|x|<1/L and diverges when ∣x∣>1/L|x|>1/L, which is precisely the Cauchy–Hadamard definition of the radius of convergence R=1/LR=1/L (with the conventions R=∞R=\infty if L=0L=0 and R=0R=0 if L=∞L=\infty). This is the most important application of the Root Test and the direct bridge to the theory of Power Series and Radius of Convergence.

Before this

Where this leads