Roselyn Daniel

2022-07-21

Determining $\underset{n\to \mathrm{\infty}}{lim}{({n}^{{\textstyle \frac{1}{n}}}-1)}^{n}$ with only elementary math

I am trying to find this limit:

$\underset{n\to \mathrm{\infty}}{lim}{({n}^{{\textstyle \frac{1}{n}}}-1)}^{n},$

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

I am trying to find this limit:

$\underset{n\to \mathrm{\infty}}{lim}{({n}^{{\textstyle \frac{1}{n}}}-1)}^{n},$

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

Beginner2022-07-22Added 11 answers

For $n\ge 5$ we have, using the binomial theorem

${(1+\frac{1}{2})}^{n}\ge 1+\frac{n}{2}+\frac{n(n-1)}{8}\ge 1+\frac{n}{2}+\frac{n(5-1)}{8}=1+n>n$

Thus $1+\frac{1}{2}\ge {n}^{1/n}$, or

$0\le ({n}^{1/n}-1{)}^{n}\le \frac{1}{{2}^{n}}$

(the first inequality being immediate.) Tking the limit as $n$ tend to $\mathrm{\infty}$, we get

$\underset{n\to \mathrm{\infty}}{lim}({n}^{1/n}-1{)}^{n}=0.$

${(1+\frac{1}{2})}^{n}\ge 1+\frac{n}{2}+\frac{n(n-1)}{8}\ge 1+\frac{n}{2}+\frac{n(5-1)}{8}=1+n>n$

Thus $1+\frac{1}{2}\ge {n}^{1/n}$, or

$0\le ({n}^{1/n}-1{)}^{n}\le \frac{1}{{2}^{n}}$

(the first inequality being immediate.) Tking the limit as $n$ tend to $\mathrm{\infty}$, we get

$\underset{n\to \mathrm{\infty}}{lim}({n}^{1/n}-1{)}^{n}=0.$

anudoneddbv

Beginner2022-07-23Added 2 answers

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).

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