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Prove that for all $p \in \mathbb{N}$ we have $\displaystyle\lim_{n\rightarrow\infty} \Large{\sqrt[n+p]{n}}=\large{1}$.

Is my demo below well defined or does I need to add something?

$$\forall \hspace{0.1cm} p \in \mathbb{N}$$ we have $$\displaystyle\lim_{n\rightarrow\infty} \Large{\sqrt[n+p]{n}}=\large{1},$$ because is valid $$ 1\ \leq\ \Large{\sqrt[n+p]{n}}\ \normalsize{\leq}\ \Large{\sqrt[n]{n}}. $$ by the squeeze theorem, we have $$\displaystyle\lim_{n\rightarrow\infty} \Large{\sqrt[n+p]{n}}=\large{1}$$ $\square $

Calvin Khor
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Fernando Sousa
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    The proof is fine, as long as you already know $\lim_{n\to\infty}\sqrt[n]{n}=1$. – J.G. May 20 '21 at 20:55
  • @J.G. How could I improve it if I didn’t know $\lim_{n\to\infty}\sqrt[n]{n}=1$? Can you Help-me? – Fernando Sousa May 20 '21 at 20:58
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    The best way would still be to prove that result first (e.g. with this), then do what you've done. – J.G. May 20 '21 at 21:06
  • I don't think the statement $1 \le \sqrt[n+p]{n}$ is valid as written. Consider $n=0.9$ and $p=1$. I think you may have to specify that $n\ge 1$. – BSplitter May 20 '21 at 21:20
  • @J.G. Thank you so much. I used Bernoulli’s inequality to show that $\lim_{n\to\infty}\sqrt[n]{n}=1$ and then continue demonstrating that I showed above. – Fernando Sousa May 20 '21 at 22:09

3 Answers3

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Your proof is fine, but in the comments you ask for a proof of the RHS, so here goes: $$\lim_{n\to \infty} n^{\frac 1n}=e^{\lim_{n\to \infty} \frac {\ln n}{n}}$$ Now this $\lim_{n\to \infty} \frac {\ln n}{n}$ can be easily evaluated using L'Hôspital rule to give limiting value $0$. Hence our net limit value is $e^0=1$.

Ritam_Dasgupta
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You can use $\lim\limits_{x\to\infty}\dfrac{\ln(x)}x=0$

Indeed for $x>0$ and $f(x)=\frac{\ln(x)}x$ then $f'(x)=\dfrac{1-\ln(x)}{x^2}<0$ for $x\gg 1$

This means that $f \searrow$ at infinity, and since the function is continuous on $[1,+\infty)$ and positive it is bounded on this interval.

Therefore $f(x)=\dfrac{\ln(x)}{x}=\dfrac{2\ln(\sqrt{x})}x=\underbrace{\dfrac 2{\sqrt{x}}}_{\to 0}\times \underbrace{f(\sqrt{x})}_\text{bounded}\to 0$

The result $\sqrt[n]{n}=\exp(\frac{\ln(n)}n)\to e^0=1$ is then immediate, and you can apply the squeeze theorem as you did to conclude.

zwim
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Since $1 < n^{1/(n+p)} < n^{1/n} $ (since $n^n < n^{n+p}$), this follows from $n^{1/n} \to 1$.

To show that, by Bernoulli's inequality, $(1+n^{-1/2})^n \ge 1+n/n^{1/2} \gt n^{1/2} $. Raising to the $2/n$ power, $n^{1/n} \lt (1+n^{-1/2})^2 =1+2n^{-1/2}+n \lt 1+3n^{-1/2} $.

For a better bound, if integer $m \ge 2$, $(1+n^{1/m-1})^n \ge 1+n^{1/m} \gt n^{1/m} $. Raising to the $m/n$ power,

$\begin{array}\\ n^{1/n} &\lt (1+n^{1/m-1})^{m}\\ &=\sum_{k=0}^m \binom{m}{k}n^{k(1/m-1)}\\ &=1+\sum_{k=1}^m \binom{m}{k}n^{k(1/m-1)}\\ &\le 1+\sum_{k=1}^m \binom{m}{k}n^{1/m-1}\\ &\le 1+(2^m-1)n^{1/m-1}\\ \end{array} $

Therefore $n^{1/n}-1 =O_m(n^{1/m-1}) $ for all integer $m \ge 2$, where $O_m$ means that the constant implied by $"O"$ depends on $m$.

For example, if $m=4$ this is $n^{1/n} \lt 1+15n^{-3/4}$.

More involved analysis can get greatly improved constants. I can show by elementary means that if $n > m^{2m/(m-1)}$ then $n^{1/n} < 1+(m^2/(m-1))n^{-1+1/m} $.

For $m=4$ this is $n^{1/n} < 1+(16/3)n^{-3/4} $ for $n > 4^{8/3} \approx 40.3$.

marty cohen
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