Elementary proof for $\lim_{n\to \infty}\sqrt[2n-1]{\dfrac{1}{2^n}}=\dfrac1{\sqrt 2}$

Question

This question is from Rudin PMA Ch 2:

How to show that $\lim_{n\to \infty}\sqrt[2n-1]{\dfrac{1}{2^n}}=\dfrac1{\sqrt 2}$?

It was pointed out to me that $\lim_{n\to\infty}\dfrac1{2^{\frac n{2n-1}}}=\dfrac1{2^{\lim_{n\to\infty}\frac n{2n-1}}}$, but I got to know that 'taking limit to power' depends on continuity. Rudin has yet to touch continuity.

Can we use the fact that $\lim_{n\to \infty}\sqrt[2n]{\dfrac{1}{2^n}}=\dfrac1{\sqrt 2}$ for this proof?

Answer 1

Yes we can use it writing

$$\sqrt[2n-1]{\dfrac{1}{2^n}} =\frac{1}{2^{\frac{n-\frac12}{2n-1}}\cdot2^{\frac{\frac12}{2n-1}}} = \sqrt[2n-1]{\dfrac{1}{2^{n-\frac12}}}\cdot \sqrt[2n-1]{\dfrac{1}{\sqrt2}}$$

but it doesn't seem to be the best method.

Answer 2

You can use the fact that $\lim_{n\to\infty}\sqrt[2n]{\frac{1}{2^n}} = \frac{1}{\sqrt 2}$ as follows. For each $n$, let $\delta_n > 0$ be such that $$ \delta_n + \frac{n}{2n} = \frac{n}{2n-1}. $$ Note that $\lim_{n\to\infty}\delta_n = \lim_{n\to\infty}\frac{1}{2(2n-1)}= 0$ by the Archimedean property of $\Bbb R$ and that $\lim_{n\to\infty}2^{-\delta_n}=\sup_{r<0} 2^r = 2^0 = 1$ where the sup is over all rationals $r < 0$. Note that this doesn't use continuity, just properties of suprema and the fact that $\delta_n\to 0$: If $\epsilon > 0$, then for all sufficiently large $n$, we have $-\delta_n > -\epsilon$, so $\lim_{n\to\infty}2^{-\delta_n}$ satisfies an equivalent characterization of what it means to be the supremum of the set $\{2^r : r \in \Bbb Q\ \text{and}\ r<0\}$. The equality $\sup_{r<0} 2^r = 2^0 = 1$ is Exercise 6 from Chapter 1 of Rudin's book. Hence, \begin{align*} \lim_{n\to\infty}\sqrt[2n-1]{\frac{1}{2^n}} = \lim_{n\to\infty}\sqrt[2n]{\frac{1}{2^n}} \cdot \frac{1}{2^{\delta_n}} = \lim_{n\to\infty}\sqrt[2n]{\frac{1}{2^n}}\cdot \lim_{n\to\infty}\frac{1}{2^{\delta_n}} = \frac{1}{\sqrt 2}\cdot 1 = \frac{1}{\sqrt 2}, \end{align*} as desired.

Answer 3

With $m=2n-1$, the expression is

$$\left(2^{-(m+1)/2}\right)^{1/m}=2^{-1/2}2^{-1/2m}.$$

Clearly the second factor tends to $1$.

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If this needs to be proven, let $2^{1/2m}=1+\epsilon>1$. Then by the binomial theorem,

$$(1+\epsilon)^{2m}=2=1+2m\epsilon+\cdots$$ (all terms are positive) and $2m\epsilon<1$.

Hence

$$1<2^{1/2m}<1+\frac1{2m}$$ which squeezes to $1$.