Elementary proof of $\lim_{n\to\infty}n(\sqrt[n]{n}-1)=\infty$

Question

This question is closely related to this question, but I am not happy with the answers there for several reasons which I will explain in a second.

The limit $\lim_{n\to\infty}n(\sqrt[n]{n}-1)=\infty$, where $n$ is a natural number, is easy to see by expanding the left side with the help of the exponential series. Indeed, we have $$ n(\sqrt[n]{n}-1)=\ln(n)+\frac{1}{2}\cdot\frac{1}{n}\cdot\ln(n)^2+\frac{1}{6}\cdot\frac{1}{n^2}\cdot\ln(n)^3+\cdots\geq\ln(n)\,. $$ Since $\ln(n)$ grows arbitrary large with $n$ large, the limit is proven.

I found this limit as an exercise in Analysis 1 by K. Königsberger, 5.8 Exercises, 3(b). I am using an old printing and the numbering might have changed, but it is in the very beginning of the book in a chapter about sequences.

At that stage of the book the exponential series as well as logarithms have not yet been introduced and very few means are available. For educational purposes, I am looking for a really elementary proof which uses a different bound from below which in turn goes to infinity. The book suggests that such a proof must exist but I cannot find one.

Can you please help me to find such proof? What is available at this stage is the Bernoulli inequality and the expansion of $(1+x)^n$ for a natural number $n$ and arbitrary $x$, plus some very basic limits like $\sqrt[n]{n}\to 1$, etc. which all can be done elementary. Thank you for your time and help!

Answer 1

For each $M \in \mathbb{R}$, choose positive integers $k$ and $N$ so that $\frac{k}{2} > M$ and $N = 2^k$. Then for each $n \geq N$, we have

\begin{align*} n \bigl( n^{1/n} - 1 \bigr) &\geq n \bigl( 2^{k/n} - 1 \bigr) \\ &= n \cdot \frac{2^{k/n} - 1}{2^{n/n} - 1} \\ &= n \cdot \frac{1 + 2^{1/n} + \cdots + 2^{(k-1)/n}}{1 + 2^{1/n} + \cdots + 2^{(n-1)/n}} \\ &\geq n \cdot \frac{k}{2n} = \frac{k}{2} > M \end{align*}

and therefore the sequence diverges to $+\infty$.

---

Remark. Note that the above argument essentially proves the inequality

$$ n \bigl( n^{1/n} - 1 \bigr) \geq \frac{1}{2}\lfloor \log_2 n \rfloor, $$

so we can't really avoid the logarithm from entering this picture, as @Thomas Andrews anticipated.

Answer 2

For $n,x\in\mathbb{Z}$, $n\ge1$ and $x\ge0$, $$ \left(1+\frac{x}{n}\right)^{n+x}\ge\left(1+\frac{x}{n+1}\right)^{n+x+1}\tag1 $$ That is, $\left(1+\frac{x}{n}\right)^{n+x}$ is non-increasing in $n$. A proof of $(1)$, using only Bernoulli's Inequality, is given below.

Suppose that $$ n\left(n^{1/n}-1\right)\le x\tag2 $$ Then, $$ \begin{align} n &\le\left(1+\frac{x}{n}\right)^n\tag{3a}\\ &\le\left(1+\frac{x}{n}\right)^{n+x}\tag{3b}\\[3pt] &\le(1+x)^{1+x}\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: follows directly from $(2)$
$\phantom{\text{(3a):}}$ (divide by $n$, add $1$, and raise to the $n^\text{th}$ power)
$\text{(3b)}$: multiply by $\left(1+\frac{x}{n}\right)^x\ge1$
$\text{(3c)}$: apply $(1)$ ($n-1$ times)

Thus, we have shown that $$ n\left(n^{1/n}-1\right)\le x\implies n\le(1+x)^{1+x}\tag4 $$ The contrapositive of $(4)$, which is equivalent to $(4)$, is $$ n\gt(1+x)^{1+x}\implies n\left(n^{1/n}-1\right)\gt x\tag5 $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}n\left(n^{1/n}-1\right)=\infty}\tag6 $$

---

Proof of $\bf{(1)}$ (repurposed from $(2)$ in this answer)

For $n,x\in\mathbb{Z}$, $n\ge1$ and $x\ge0$, $$ \begin{align} \frac{\left(1+\frac{x}{n}\right)^{n+x}}{\left(1+\frac{x}{n+1}\right)^{n+x+1}} &=\left(\frac{n+x}{n}\right)^{n+x}\left(\frac{n+1}{n+x+1}\right)^{n+x+1}\tag{7a}\\ &=\frac{n}{n+x}\left(\frac{n+x}{n}\frac{n+1}{n+x+1}\right)^{n+x+1}\tag{7b}\\ &=\frac{n}{n+x}\left(\frac{n^2+(x+1)n+x}{n^2+(x+1)n}\right)^{n+x+1}\tag{7c}\\ &=\frac{n}{n+x}\left(1+\frac{x}{(n+x+1)n}\right)^{n+x+1}\tag{7d}\\[3pt] &\ge\frac{n}{n+x}\left(1+\frac{x}{n}\right)\tag{7e}\\[6pt] &=1\tag{7f} \end{align} $$ Explanation:
$\text{(7a)}$-$\text{(7d)}$: algebraic manipulation
$\text{(7e)}$: Bernoulli's Inequality
$\text{(7f)}$: simplification

Answer 3

By way of contradiction suppose it does not converge to $\infty$. Then there exists a subsequence $n_k$ with $\lim_{k\rightarrow \infty} n_k = \infty$ s.t. $$\lim_{k \rightarrow \infty} n_k\left(n_k^{1/n_k}-1\right) < \infty$$ i.e. $$n_k\left(n_k^{1/n_k}-1\right) < c \\ n_k < \left(1+\frac{c}{n_k}\right)^{n_k}$$ $\forall k\in \mathbb{N}$ and some $c>0$.

Then use $$\left(1+x/n\right)^n = \sum_{k=0}^n \binom{n}{k} \left(\frac{x}{n}\right)^k = \sum_{k=0}^n \frac{n}{n}\frac{(n-1)}{n}\cdots\frac{(n-k+1)}{n} \, \frac{x^k}{k!} \leq \sum_{k=0}^n \frac{x^k}{k!} \leq \sum_{k=0}^\infty \frac{x^k}{k!}$$ and the RHS clearly converges since $$\lim_{k\rightarrow \infty} \frac{x^{k+1}/(k+1)!}{x^k/k!} = \lim_{k\rightarrow \infty}\frac{x}{k+1}=0<1 \,.$$

Answer 4

You can show it converges to infinity if you known:

Lemma: For any real $C>0,$ the sequence $$\left(1+\frac{C}n\right)^n$$ is bounded above.

Given this lemma, we can prove your result, because:

$$\begin{align}n(\sqrt[n]{n}-1)&>C\iff \\n\sqrt[n]{n}&> n+C\iff\\n^{n+1}&>(C+n)^{n}\iff\\ n&>\left(1+\frac Cn\right)^n\end{align}$$

By the lemma, when $n$ is large enough, this last inequality is true.

Proof of lemma:

We use that:

$$\binom{m}{k}\leq\frac{m^k}{k!}\tag 1$$ and, for $n\geq m,$ $$n!\geq m^{n-m} m!.\tag2$$ These two are equivalent, and easily proven.

Then we choose some $n_0>C.$

For $n\geq n_0,$ we have: $$\begin{align}\left(1+\frac Cn\right)^{n}&=\sum_{k=0}^{n}\binom nk\frac{C^k}{n^k}\\&\leq\sum_{k=0}^n \frac{C^k}{k!}\\&\leq \sum_{k=0}^{n_0-1}\frac{C^k}{k!}+\sum_{k=n_0}^\infty\frac{C^k}{n_0!n_0^{k-n_0}}\\ &=\sum_{k=0}^{n_0-1} \frac{C^k}{k!}+\frac{C^{n_0}}{n_0!}\frac{1}{1-\frac{C}{n_0}} \end{align}$$

---

Note, although this answer is essentially proving the power series for $e^C$ converges, we don't need to know that this power series is the limit, nor anything about the function $e^x.$

Nor do we need that $(1+C/n)^n$ converges.

Answer 5

You only need the $\text{HM}\leq\text{GM}\leq\text{AM}$ inequality and few other ingredients (harmonic numbers).
For any natural number $n\geq 2$ we have $$ n = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) \tag{TelescopicProduct}$$ hence $$ \sqrt[n]{n}=GM\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right) $$ is upper bounded by $$AM\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)=1+\frac{H_{n-1}}{n} $$ and lower bounded by $$HM\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)=\frac{n}{n+1-H_n}$$ hence $$ \sqrt[n]{n}-1 \geq \frac{(H_n-1)}{n-(H_n-1)}\geq \frac{H_n-1}{n} $$ and $n(\sqrt[n]{n}-1)$ is clearly divergent.

Answer 6

Let $n^{1/n}=1+d_n.$ If $nd_n\not\to\infty$ then there exists $k>0$ such that $d_n<k/n$ for infinitely many $n.$ If $d_n<k/n$ then by the Binomial Theorem $$n=(1+d_n)^n=\sum_{j=0}^n(d_n)^j\binom n j \le$$ $$\le\sum_{j=0}^n(k/n)^j\binom n j\le$$ $$\le\sum_{j=0}^nk^j/ j!<$$ $$<\sum_{j=0}^ {\infty}k^j/j!.$$ But the last series (above) is convergent because when $j>2k$ the ratio of successive terms $\frac { k^{j+1}/ (j+1)!}{k^j/ j!}$ is less than $1/2.$

So if $nd_n\not\to\infty$ then there are infinitely many $n\in \Bbb N$ that are less than the real number $\sum_{j=0}^{\infty}k^j/ j!,$ which is absurd.