Prove that $c^{1/n} \rightarrow 1$ for $c > 0$. I can show a straightforward $\epsilon-\delta$ proof that when $c > 1$, $c^{1/n} \rightarrow 1$. But not when $c < 1$?
WTS
For all $\epsilon$ there exists $N$ st. for $n \geq N$, $|c^{1/n} - 1| < \epsilon$
When $c < 1$, the absolute value can be removed and the inner expression changed to:
$1 - c^{1/n}$. We can temporarily set this $< \epsilon$, and see if a useful expression for $n$ in terms of $\epsilon$ comes out:
$1 - c^{1/n} < \epsilon$
$1 - \epsilon < c^{1/n}$
$\log_{c}(1 - \epsilon) < \frac{1}{n}$
$log_{1 - \epsilon}(c) > n$.
The final expression is not terms of $\text{"something"} < n$, which is usually expected. What might be wrong with what I showed?

(very unoriginal)
By Bernoulli's inequality,
$(1+a/n)^n
\ge 1+a
$
so
$(1+a)^{1/n}
\le 1+a/n
$.
If $c > 1$,
let
$c = 1+a$.
Then
$c^{1/n}
= (1+a)^{1/n}
\le 1+a/n
\to 1
$
as
$n \to \infty$.
If
$0 < c < 1$,
let
$c^{1/n} = \dfrac1{1+b}
$.
Then
$c
= \dfrac1{(1+b)^n}
\le \dfrac1{1+nb}
\lt \dfrac1{nb}
$
so
$b < \dfrac1{nc}
$
and
$c^{1/n} 
= \dfrac1{1+b}
\gt \dfrac1{1+\dfrac1{nc}}
= \dfrac{nc}{nc+1}
= 1-\dfrac{1}{nc+1}
\to 1
$
as $n \to \infty$.
From
"What is Mathematics"
by Courant and Robbins.
Read this book!
