LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$n^{1 / n} \to 1$

Statement

For a natural number $n \geq 1$ write $ι (n) : = n \cdot 1_{R}$ for the canonical natural of $R$ (Canonical naturals are positive and strictly increasing) and $n^{1 / n} : = ι (n)^{1 / n}$ , $n^{1 / 2} : = ι (n)^{1 / 2}$ for its roots (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base). Then:

$1 \leq n^{1 / n} \leq 1 + \frac{2}{n^{1 / 2}}$ for every natural $n \geq 1$ ;
the sequence $r_{k} : = (k + 1)^{1 / (k + 1)}$ , $k \in N$ , converges to $1$ (Limits and Cauchy sequences of reals).

The index range is not cosmetic. The expression $n^{1 / n}$ is defined only for $n \geq 1$ , since $1 / n$ is not a rational number when $n = 0$ (Rational powers $a^{r}$ of a positive base). Sequences in this library are functions on $N$ and $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so the statement of convergence is made about the shifted family $r_{k} = (k + 1)^{1 / (k + 1)}$ , which is the classical family $n^{1 / n}$ , $n \geq 1$ , reindexed by $n = k + 1$ . Claim 1 is stated over the natural range $n \geq 1$ where the expression means something.

Facts & Assumptions

Given: For a natural $m \geq 1$ the canonical natural $ι (m) : = m \cdot 1_{R}$ , extended by $ι (0) : = 0$ ; this extension keeps the additivity $ι (m + m^{'}) = ι (m) + ι (m^{'})$ of Canonical naturals are positive and strictly increasing, which for $m$ or $m^{'}$ equal to $0$ reads $ι (m) = ι (m) + 0$ .

[L1]

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

[L2]

Rational powers and monotonicity: $a^{1 / n}$ is the rational power $a^{r}$ at $r = 1 / n$ , and for rational $t > 0$ one has $a^{t} > 1$ whenever $a > 1$ ; also $(1 / a)^{1 / 2} = 1 / a^{1 / 2}$ for $a > 0$ (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]

AM-GM: for a natural $n \geq 1$ and reals $a_{0}, \dots, a_{n - 1} \geq 0$ , the geometric mean $(\prod_{j < n} a_{j})^{1 / n}$ is $\leq$ the arithmetic mean $\frac{1}{ι (n)} \sum_{j < n} a_{j}$ (The arithmetic mean, geometric mean inequality).

[L4]

Finite sums and products: the empty sum is $0$ and the empty product $1$ ; sums and products split at any intermediate index; and $\sum_{j < m} λ = ι (m) λ$ for a constant $λ$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Induction principle (The principle of mathematical induction).

[L6]

Canonical naturals: $ι (m) > 0$ and $ι (m)$ is invertible for $m \geq 1$ , $ι$ is strictly increasing, and $ι (2) = 2 > 1$ ; the Archimedean property gives, for every real $x$ , a natural $p \geq 1$ with $x < ι (p)$ (Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean).

[L7]

Order and reciprocals: $0 < a < b$ gives $0 < 1 / b < 1 / a$ ; multiplying an inequality by a positive element preserves it; and inequalities may be added and translated (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities).

[L8]

Squares: for $a, b \geq 0$ one has $a < b$ if and only if $a \cdot a < b \cdot b$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Integer powers $a^{m}$ ).

[L9]

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

[L10]

The order on $N$ is total and $ι$ respects it (Order on the natural numbers, $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1givenL1L6

For a natural $n \geq 1$ the element $ι (n)$ is positive and invertible, so $ι (n)^{1 / n}$ and $ι (n)^{1 / 2}$ exist and are positive.

1.2givenL4L5

For every natural $m$ one has $\prod_{j < m} 1 = 1$ : the empty product is $1$ , and if $\prod_{j < m} 1 = 1$ then $\prod_{j < m + 1} 1 = (\prod_{j < m} 1) \cdot 1 = 1$ , so this follows by induction on $m$ .

2.1step 1.1L1L2L6L10

For $n = 1$ one has $ι (1) = 1$ and $1^{1 / 1} = 1$ ; for $n \geq 2$ one has $ι (n) \geq ι (2) = 2 > 1$ and $1 / n$ is a positive rational, so $ι (n)^{1 / n} > 1$ . In either case $n^{1 / n} \geq 1$ .

2.2step 1.1step 1.2L1L3L4L6L7algebra

Let $n \geq 2$ and put $u : = ι (n)^{1 / 2}$ , so that $u > 0$ and $u \cdot u = ι (n)$ . Apply [L3] to the list of $n$ nonnegative reals given by $a_{0} = a_{1} = u$ and $a_{j} = 1$ for $2 \leq j < n$ , the latter range being empty when $n = 2$ . Splitting at index $2$ gives $\prod_{j < n} a_{j} = (\prod_{j < 2} a_{j}) (\prod_{j < n - 2} a_{2 + j}) = (u \cdot u) \cdot 1 = ι (n)$ by step 1.2, so the geometric mean is $ι (n)^{1 / n}$ ; and $\sum_{j < n} a_{j} = (\sum_{j < 2} a_{j}) + (\sum_{j < n - 2} 1) = (u + u) + ι (n - 2) = (u + u) + ι (n) - 2$ , using additivity of $ι$ and $ι (2) = 2$ , so the arithmetic mean is $A = ((u + u) + ι (n) - 2) / ι (n) = 1 + ((u + u) - 2) / ι (n)$ . Since $(u + u) - 2 < u + u$ and $ι (n) > 0$ , and $(u + u) / ι (n) = (u + u) / (u \cdot u) = 2 / u$ , this gives $ι (n)^{1 / n} \leq A \leq 1 + 2 / u = 1 + 2 / n^{1 / 2}$ .

2.3step 1.1L1L6L7

For $n = 1$ the same bound holds trivially: $1^{1 / 1} = 1 \leq 1 + 2 = 1 + 2 / 1^{1 / 2}$ .

2.4step 1.1L1L6L7L8L9L10algebra

The sequence $b_{k} : = 1 + 2 / (k + 1)^{1 / 2}$ converges to $1$ . Given a real $ε > 0$ , put $t : = 2 / ε > 0$ and take a natural $p \geq 1$ with $t \cdot t < ι (p)$ . For $k \geq p$ we have $k + 1 > p$ , hence $ι (k + 1) > ι (p) > t \cdot t$ , and since $(ι (k + 1)^{1 / 2}) (ι (k + 1)^{1 / 2}) = ι (k + 1)$ with both factors $\geq 0$ , this forces $t < ι (k + 1)^{1 / 2}$ . Therefore $0 < 2 / ι (k + 1)^{1 / 2} < 2 / t = ε$ , that is $∣ b_{k} - 1 ∣ < ε$ .

3.1step 2.1step 2.2step 2.3

Claim 1 is the combination of steps 2.1, 2.2 and 2.3, the two upper bounds covering $n \geq 2$ and $n = 1$ respectively.

4.1step 3.1step 2.4L9∎

For every $k \in N$ the natural $k + 1$ is $\geq 1$ , so claim 1 gives $1 \leq r_{k} \leq b_{k}$ . The constant sequence $1$ converges to $1$ and $(b_{k})$ converges to $1$ by step 2.4, so the squeeze theorem gives $r_{k} \to 1$ , which is claim 2.

Remarks

Where the $\sqrt{n}$ comes from. AM-GM is applied to a list whose product is $n$ but whose entries are as close to $1$ as possible: two copies of $n^{1 / 2}$ and $n - 2$ copies of $1$ . The arithmetic mean is then $1 + (2 n^{1 / 2} - 2) / n$ , which tends to $1$ at the rate $2 / n^{1 / 2}$ . Splitting $n$ as $n^{1 / 2} \cdot n^{1 / 2}$ rather than as $n \cdot 1$ is the whole trick: the list $n, 1, \dots, 1$ gives only $n^{1 / n} \leq 2 - 1 / n$ , which does not converge to $1$ .
The lower bound is not decoration. Without $n^{1 / n} \geq 1$ the squeeze has nothing below it, and the upper bound alone would leave open a limit smaller than $1$ . It comes from monotonicity of rational powers in the base (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ) and holds with equality only at $n = 1$ .
No logarithm and no exponential is used. The usual quick proof writes $n^{1 / n} = e^{(\log n) / n}$ and appeals to $(\log n) / n \to 0$ ; neither function exists in this library yet, and the AM-GM route needs nothing beyond roots and finite sums.

$n^{1 / n} \to 1$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

n1/n→1n1/n→1

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

$n^{1 / n} \to 1$