Prove that for all $\ p \in \mathbb{N}\ $ we have $\ \displaystyle\lim_{n\rightarrow\infty} {\sqrt[n+p]{n}}={1}$ 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 $
 A: 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 $e^0=1$.
A: 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.
A: 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
$n > 4^{8/3} \approx 40.3$.
