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

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For every $p > 0$ and every positive rational $\alpha$ , $n^{\alpha}/(1+p)^n \to 0$

Statement

Let $p \in \mathbb{R}$ with $p > 0$ and let $\alpha \in \mathbb{Q}$ with $\alpha > 0$ . Write $\iota(n) := n \cdot 1_{\mathbb{R}}$ for the canonical natural, with $\iota(0) := 0$ , and let

$w_k \;:=\; \frac{\iota(k)^{\alpha}}{(1+p)^{k}} \qquad (k \in \mathbb{N}),$

the numerator being a rational power (Rational powers $a^r$ of a positive base) and the denominator an integer power (Integer powers $a^m$ ). Then $w_k \to 0$ (Limits and Cauchy sequences of reals).

Every term is defined, including the one at $k = 0$ . The supplementary clause of Rational powers $a^r$ of a positive base gives $0^{\alpha} = 0$ for rational $\alpha > 0$ , and $(1+p)^0 = 1$ , so $w_0 = 0$ . No index shift is therefore needed here, in contrast with the two root lemmas earlier on this page, where the exponent is the index.

In words: a fixed power of $n$ is beaten by any geometric sequence of ratio $> 1$ , however small the excess $p$ and however large the exponent $\alpha$ .

Facts & Assumptions

Given: A real $p > 0$ and a rational $\alpha > 0$ ; the base $\beta := 1 + p > 1$ ; the canonical naturals $\iota(n) = n \cdot 1_{\mathbb{R}}$ with $\iota(0) = 0$ ; and $w_k = \iota(k)^{\alpha}/\beta^{k}$ .

[L1]

Rational powers: $x^r$ is defined and positive for real $x > 0$ and rational $r$ , and $0^{r} = 0$ for rational $r > 0$ ; the integer power $x^{m}$ is the rational power at exponent $m$ ; $(xy)^{r} = x^{r} y^{r}$ , which persists for $x, y \ge 0$ when $r > 0$ ; $x^{-r} = 1/x^{r}$ ; and $(x^{r})^{s} = x^{rs}$ (Rational powers $a^r$ of a positive base, Laws of rational exponents, Integer powers $a^m$ , Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ ).

[L2]

Monotonicity of rational powers: for rational $t > 0$ , $x > 1$ implies $x^{t} > 1$ ; and for rational $t > 0$ , $0 < x < y$ implies $x^{t} < y^{t}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L3]

Integer powers: $x > 0$ implies $x^{m} > 0$ , and $x^{m} x^{m'} = x^{m+m'}$ , $(x^{m})^{m'} = x^{m m'}$ for integer exponents with $x \ne 0$ (Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ , Laws of integer exponents).

[L4]

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

[L5]

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 $\mathbb{N}$ ).

[L6]

Reciprocal Archimedean property: for every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ ; 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).

[L7]

Order arithmetic: Order is preserved by adding a constant and by adding inequalities and claim 4 of Sign rules for products and monotonicity of multiplication state the strict forms, that inequalities may be translated and added and that multiplication by a positive element preserves $<$ ; adjoining the case of equality gives the nonstrict forms used below. Products of nonnegative inequalities multiply in the nonstrict form stated by Multiplying inequalities of positives, and the order is total (Ordered field).

[L8]

Convergence to $0$ : it suffices to produce, for every real $\varepsilon > 0$ , a threshold beyond which $|z_k| < \varepsilon$ ; and $|z| = z$ for $z \ge 0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Basic properties of the absolute value).

Proof

technique · direct

1.1

Since $\alpha > 0$ is rational, so is $1/\alpha$ ; put $\delta := \beta^{1/\alpha}$ and $\theta := \delta^{1/2}$ . From $\beta > 1$ and $1/\alpha > 0$ we get $\delta > 1$ , and from $\delta > 1$ and $1/2 > 0$ we get $\theta > 1$ ; hence $\theta - 1 > 0$ and $\delta > 0$ , $\theta > 0$ .

givenL1L2L7

1.2

For every natural $n$ one has $\delta^{n} = \theta^{n} \theta^{n}$ , because $\theta^{2} = (\delta^{1/2})^{2} = \delta$ and therefore $\delta^{n} = (\theta^{2})^{n} = \theta^{2n} = \theta^{n} \theta^{n}$ .

givenL1L3

1.3

For every natural $k$ one has $w_k = u_k^{\alpha}$ , where $u_k := \iota(k)/\delta^{k}$ . Indeed $u_k = \iota(k) \cdot (1/\delta^{k})$ with both factors $\ge 0$ , so $u_k^{\alpha} = \iota(k)^{\alpha} \big(1/\delta^{k}\big)^{\alpha} = \iota(k)^{\alpha}/\big(\delta^{k}\big)^{\alpha}$ , and $\big(\delta^{k}\big)^{\alpha} = \delta^{k\alpha} = \big(\beta^{1/\alpha}\big)^{k\alpha} = \beta^{(1/\alpha)(k\alpha)} = \beta^{k}$ .

givenL1L3

2.1

For every natural $n \ge 1$ one has $0 \le u_n < 1/\big(\iota(n)(\theta-1)(\theta-1)\big)$ . Bernoulli's inequality applied to $\theta - 1 > 0$ gives $\theta^{n} \ge 1 + \iota(n)(\theta-1) > \iota(n)(\theta-1) > 0$ , so multiplying this inequality by itself gives $\delta^{n} = \theta^{n}\theta^{n} > \iota(n)(\theta-1)\iota(n)(\theta-1) > 0$ ; dividing the positive $\iota(n)$ by the two positive quantities reverses the inequality and yields $u_n = \iota(n)/\delta^{n} < \iota(n)/\big(\iota(n)\iota(n)(\theta-1)(\theta-1)\big) = 1/\big(\iota(n)(\theta-1)(\theta-1)\big)$ , while $u_n \ge 0$ because $\iota(n) > 0$ and $\delta^{n} > 0$ .

step 1.1step 1.2L3L4L5L6L7

3.1

The sequence $(u_k)$ converges to $0$ . Note first $u_0 = \iota(0)/\delta^{0} = 0/1 = 0$ . Given a real $\varepsilon > 0$ , put $\eta := \varepsilon(\theta-1)(\theta-1) > 0$ and take a natural $m \ge 1$ with $1/m < \eta$ . For $k \ge m$ we have $\iota(k) \ge \iota(m) > 0$ , hence $1/\iota(k) \le 1/\iota(m) < \eta$ , and therefore $0 \le u_k < 1/\big(\iota(k)(\theta-1)(\theta-1)\big) < \eta/\big((\theta-1)(\theta-1)\big) = \varepsilon$ , so $|u_k| < \varepsilon$ .

step 2.1L5L6L7L8

4.1

The sequence $(w_k)$ converges to $0$ . Given a real $\varepsilon > 0$ , the element $\varepsilon^{1/\alpha}$ is a positive real, so by step 3.1 there is a threshold beyond which $0 \le u_k < \varepsilon^{1/\alpha}$ . For such $k$ : if $u_k = 0$ then $w_k = 0^{\alpha} = 0 < \varepsilon$ , and if $u_k > 0$ then monotonicity of the rational power $\alpha$ in the base gives $w_k = u_k^{\alpha} < \big(\varepsilon^{1/\alpha}\big)^{\alpha} = \varepsilon^{(1/\alpha)\alpha} = \varepsilon$ . In both cases $|w_k| = w_k < \varepsilon$ , so $w_k \to 0$ .

step 3.1step 1.3L1L2L8∎

Remarks

The proof turns the problem into the single case $\alpha = 1$ . Writing $\beta = \delta^{\alpha}$ with $\delta := \beta^{1/\alpha}$ makes $w_k = \big(\iota(k)/\delta^{k}\big)^{\alpha}$ , so it is enough to know that $n/\delta^{n} \to 0$ for a base $\delta > 1$ , and then that a fixed positive rational power of a nonnegative null sequence is null. The exponent never has to be moved inside a limit.
Bernoulli is applied to the square root of the base, and that is essential. Applied to $\delta$ itself it gives only $\delta^{n} \ge 1 + n(\delta-1)$ , which makes $n/\delta^{n}$ bounded but not null. Applied to $\theta = \delta^{1/2}$ and then squared it gives $\delta^{n} > n^{2}(\theta-1)^{2}$ , a quadratic lower bound, and one factor of $n$ is then left over to drive the quotient to $0$ .
Why the route through the root test is not taken. The chain of For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ together with $n^{1/n} \to 1$ would give $\limsup_k w_k^{1/k} \le \limsup_k \big((k^{1/k})^{\alpha}\big)/\beta$ , and closing that requires knowing $t_k \to 1 \Rightarrow t_k^{\alpha} \to 1$ , that is the continuity of $x \mapsto x^{\alpha}$ at $x = 1$ . That statement is not available at this point in the reading order and is not proved on this page; it is proved later in Continuity and derivatives of positive-base real powers ↗, so the argument above is made directly instead. It needs only Bernoulli and the Archimedean property.
The growth hierarchy this places. Together with For every real $x$ , $x^k/k! \to 0$ it orders the three standard scales: a fixed power of $n$ is beaten by every geometric sequence of ratio $> 1$ , and every geometric sequence is beaten by $k!$ . Worked instances are collected in The four standard limits $n^{1/n} \to 1$ , $a^{1/n} \to 1$ , $n^{\alpha}/(1+p)^n \to 0$ and $x^k/k! \to 0$ , computed ↗.

Depends on

Used by

The four standard limits n^1/n → 1, a^1/n → 1, n^α/(1+p)ⁿ → 0 and xᵏ/k! → 0, computed Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 92 results over 28 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

Bernoulli's inequality (Wikipedia) (standard reference, not scraped)
Exponential growth (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.20) (standard reference, not scraped)

For every p>0p > 0p>0 and every positive rational α\alphaα, nα/(1+p)n→0n^{\alpha}/(1+p)^n \to 0nα/(1+p)n→0