Roselyn Daniel

2022-07-21

Determining $\underset{n\to \mathrm{\infty }}{lim}{\left({n}^{\frac{1}{n}}-1\right)}^{n}$ with only elementary math
I am trying to find this limit:
$\underset{n\to \mathrm{\infty }}{lim}{\left({n}^{\frac{1}{n}}-1\right)}^{n},$
I tried using exponential function, but I see no way at the moment. I am not allowed to use any kind of differentiation or other topics of advanced math, only induction and school math are possible. Thank you

eri1ti0m

For $n\ge 5$ we have, using the binomial theorem
${\left(1+\frac{1}{2}\right)}^{n}\ge 1+\frac{n}{2}+\frac{n\left(n-1\right)}{8}\ge 1+\frac{n}{2}+\frac{n\left(5-1\right)}{8}=1+n>n$
Thus $1+\frac{1}{2}\ge {n}^{1/n}$, or
$0\le \left({n}^{1/n}-1{\right)}^{n}\le \frac{1}{{2}^{n}}$
(the first inequality being immediate.) Tking the limit as $n$ tend to $\mathrm{\infty }$, we get
$\underset{n\to \mathrm{\infty }}{lim}\left({n}^{1/n}-1{\right)}^{n}=0.$

anudoneddbv

Perhaps the simplest way is to go step by step. First, show that ${n}^{1/n}\to 1$ as $n\to \mathrm{\infty }$. You can then conclude that $0<{n}^{1/n}-1<\frac{1}{2}$ for large enough values of $n$, meaning that, when $n\to \mathrm{\infty },$ the limit must be $0$ (sandwich principle).