If $c > 1$, show that $\sqrt[n]{c}$ approaches $1$ for large $n$.

Question

I know there's a duplicate about this question, but I came up with this solution on my own so I need a verification.

Using Bernoulli's inequality, we obtain $\forall \varepsilon > 0 $, $\exists N > 0$ such that if $n > N$, then $1 < 1 + n\varepsilon \le (1 + \varepsilon)^n$. However, as $n \to \infty$, $1 + n \varepsilon \to \infty$. By squeeze theorem, we have $(1 + \varepsilon)^n \to \infty$ as well.

Here comes the part I am not sure of.

For large enough $n$, we always could have a fixed $c$, such that $(1 + \varepsilon)^n > c$ $\forall \varepsilon > 0$.

This yields $c^{\frac{1}{n}} - 1 < \varepsilon$. Since $c > 1$, we have $c^{\frac{1}{n}} > 1$, so $\forall \varepsilon > 0$, $c^{\frac{1}{n}} - 1 < \varepsilon$ gives the desired result.

Is this proof lacking in any manners? Can we assume the bold sentence? Any help is greatly appreciated.

Answer 1

Your bold statement is false, I think. As stated it is difficult to parse. Notice that for any $c,n$ (large or fixed or otherwise) $$ \lim_{\epsilon\rightarrow 0} (1+\epsilon)^n=1<c $$ so the inequality never holds for all $\epsilon$. What you want is for all fixed $c>1$ and $\epsilon>0$ that $$ (1+\epsilon)^n>c $$ for all $n$ sufficiently large. From there, you get $1<c^{1/n}<1+\epsilon$ and procede as you have already done. But the statement should go with picking $c$ and $\epsilon$ before $n$ since you are taking a limit in $n$.

Answer 2

Your bold sentence is correct. Here's the proof. For any fixed $c>0$ and for any $\varepsilon>0$ we choose $\displaystyle M=\log_{(1+\varepsilon)}(c)+1$. Then any $n>M$ would do the work and vice-versa.

Answer 3

Alternative approach:

For fixed $c \in \Bbb{R_{> 1}}$ and any $n \in \Bbb{Z_{> 1}}$, binomial expansion gives that

$$\left[ ~1 + \frac{c-1}{n} ~\right]^n > c \implies $$

$$\left[ ~1 + \frac{c-1}{n} ~\right] > c^{(1/n)}.$$

Also, for fixed $c$,

$$\lim_{n\to \infty} \left[ ~1 + \frac{c-1}{n} ~\right] = 1.$$