LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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For every $a > 0$ , $a^{1/n} \to 1$

Statement

Let $a \in \mathbb{R}$ with $a > 0$ , write $\iota(n) := n \cdot 1_{\mathbb{R}}$ for the canonical natural (Canonical naturals are positive and strictly increasing) and $a^{1/n}$ for the $n$ -th root (Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ , Rational powers $a^r$ of a positive base), defined for naturals $n \ge 1$ . Then:

for every real $b \ge 1$ and every natural $n \ge 1$ , $1 \;\le\; b^{1/n} \;\le\; 1 + \frac{b-1}{\iota(n)};$
the sequence $c_k := a^{1/(k+1)}$ , $k \in \mathbb{N}$ , converges to $1$ (Limits and Cauchy sequences of reals).

Index range. As for the previous lemma on this page, $a^{1/n}$ requires $n \ge 1$ , so the sequence indexed by $\mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences) is the shifted family $a^{1/(k+1)}$ ; it is the classical family $a^{1/n}$ , $n \ge 1$ , reindexed by $n = k+1$ .

Facts & Assumptions

Given: A real $a > 0$ ; the canonical naturals $\iota(n) = n \cdot 1_{\mathbb{R}}$ for $n \ge 1$ ; and the sequence $c_k := a^{1/(k+1)}$ .

[L1]

Roots: for real $x \ge 0$ and natural $n \ge 1$ there is a unique real $s \ge 0$ with $s^n = x$ , written $x^{1/n}$ ; it is $> 0$ when $x > 0$ , and $1^{1/n} = 1$ by uniqueness (Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ , Integer powers $a^m$ ).

[L2]

Rational powers: $x^{1/n}$ is the rational power at exponent $1/n$ ; for rational $t > 0$ , $x > 1$ implies $x^t > 1$ ; and $(xy)^{1/n} = x^{1/n} y^{1/n}$ for $x, y > 0$ (Rational powers $a^r$ of a positive base, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Laws of rational exponents).

[L3]

Bernoulli's inequality: $(1+x)^n \ge 1 + \iota(n) x$ for $x \ge -1$ and $n \in \mathbb{N}$ (Bernoulli's inequality $(1+x)^n \ge 1 + nx$ ).

[L4]

Canonical naturals: $\iota(n) > 0$ and invertible for $n \ge 1$ , and $\iota$ is strictly increasing (Canonical naturals are positive and strictly increasing, Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ ).

[L5]

Reciprocal Archimedean property: for every real $\eta > 0$ there is a natural $p \ge 1$ with $1/p < \eta$ ; and $0 < x < y$ gives $0 < 1/y < 1/x$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L6]

Order arithmetic: inequalities may be added and translated, and multiplying an inequality by a positive element preserves it; the order is total, so exactly one of $a < 1$ , $a = 1$ , $a > 1$ holds (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)).

[L7]

Squeeze theorem; a constant sequence converges to its value; to establish convergence it suffices to produce a threshold for every real $\varepsilon > 0$ (The squeeze theorem, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L8]

Algebra of limits, reciprocal rule: if $z_j \to z$ with $z \ne 0$ and $z_j \ne 0$ for every $j$ , then $1/z_j \to 1/z$ (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · cases

1.1

Let $b$ be any real with $b \ge 1$ and let $n \ge 1$ be a natural. If $b = 1$ then $b^{1/n} = 1$ and both inequalities hold. If $b > 1$ then $1/n$ is a positive rational, so $t := b^{1/n} - 1 > 0$ ; Bernoulli's inequality applied to $t \ge -1$ gives $b = \big(b^{1/n}\big)^n = (1+t)^n \ge 1 + \iota(n)t$ , hence $\iota(n) t \le b - 1$ and $t \le (b-1)/\iota(n)$ since $\iota(n) > 0$ . In both cases $1 \le b^{1/n} \le 1 + (b-1)/\iota(n)$ , which is claim 1.

givenL1L2L3L4L6

1.2

Case one: $a = 1$ .

givenassume-case one

1.3

Case big: $a > 1$ .

givenassume-case big

1.4

Case small: $0 < a < 1$ .

givenassume-case small

2.1

For every real $b > 1$ the sequence $b^{1/(k+1)}$ converges to $1$ . Put $d_k := 1 + (b-1)/\iota(k+1)$ . Given a real $\varepsilon > 0$ , the quotient $\varepsilon/(b-1)$ is positive, so there is a natural $p \ge 1$ with $1/p < \varepsilon/(b-1)$ ; for $k \ge p$ we have $k+1 > p$ , hence $\iota(k+1) > \iota(p) > 0$ and $0 < (b-1)/\iota(k+1) < (b-1)(1/p) < \varepsilon$ , so $|d_k - 1| < \varepsilon$ and $d_k \to 1$ . By step 1.1 applied at $n = k+1$ we have $1 \le b^{1/(k+1)} \le d_k$ for every $k$ , and the constant sequence $1$ converges to $1$ , so the squeeze theorem gives $b^{1/(k+1)} \to 1$ .

step 1.1L4L5L6L7

2.2

In case one, $c_k = 1^{1/(k+1)} = 1$ for every $k$ , so $(c_k)$ is the constant sequence $1$ and converges to $1$ .

step 1.2L1L7

3.1

In case big, $a > 1$ , so step 2.1 applied with $b = a$ gives $c_k = a^{1/(k+1)} \to 1$ .

step 2.1step 1.3

3.2

In case small, put $a' := 1/a$ , which satisfies $a' > 1$ because $0 < a < 1$ . For each natural $n \ge 1$ the product rule for roots gives $a^{1/n} (a')^{1/n} = (a a')^{1/n} = 1^{1/n} = 1$ , so $a^{1/n} = 1/(a')^{1/n}$ , and $(a')^{1/n} > 0$ . By step 2.1 the sequence $(a')^{1/(k+1)}$ converges to $1 \ne 0$ with all terms nonzero, so the reciprocal rule gives $c_k = 1/(a')^{1/(k+1)} \to 1/1 = 1$ .

step 2.1step 1.4L1L2L5L8

4.1

The three cases are exhaustive by trichotomy applied to $a$ and $1$ , the hypothesis $a > 0$ excluding nothing else, and in each of them $(c_k)$ converges to $1$ ; together with step 1.1 this proves both claims.

step 2.2step 3.1step 3.2step 1.1L6cases: trichotomy of the ordercases-exhaustive∎

Remarks

Bernoulli is doing the whole job in the case $a > 1$ . The inequality $(1+t)^n \ge 1 + nt$ converts the exact identity $\big(a^{1/n}\big)^n = a$ into the linear bound $t \le (a-1)/n$ on the excess $t = a^{1/n} - 1$ , and that bound is what tends to $0$ . No estimate on $a^{1/n}$ itself is needed beyond $a^{1/n} > 1$ .
The case $0 < a < 1$ is not symmetric to the case $a > 1$ and is not proved again. It is transported by the reciprocal, using $a^{1/n} (1/a)^{1/n} = 1$ (Laws of rational exponents) and the reciprocal rule of Algebra of limits: sums, scalar multiples, products and quotients. The hypothesis of that rule, that the limit be nonzero and every term nonzero, is met because roots of positive reals are positive.
The rate is different from the one in $n^{1/n} \to 1$ . Here the excess is $O(1/n)$ with a constant depending on $a$ ; there the base itself grows with $n$ and the excess is only $O(1/n^{1/2})$ . The two lemmas are therefore not instances of one another in either direction.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 98 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Nth root (Wikipedia) (standard reference, not scraped)
Bernoulli's inequality (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.20) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.5 (standard reference, not scraped)
MIT 18.100B, Fall 2011, Problem Set 1 solutions (standard reference, not scraped)
N. Donaldson, Math 140A: Real Analysis notes (standard reference, not scraped)

For every a>0a > 0a>0, a1/n→1a^{1/n} \to 1a1/n→1