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What is the limit of $\displaystyle{ \lim \limits_{n \to \infty} \sqrt[n]{\sqrt[2^n]{a}-1}}$ given $a>1$ ?

I did some computations and I feel its $\frac{1}{2}$ , i don't know how to prove it

I did used Bernoulli inequality

José Carlos Santos
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Ahmad
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3 Answers3

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Note that$$\lim_{n\to\infty}n\left(\sqrt[n]a-1\right)=\log a\tag{1}.$$This equality comes from $$\lim_{n\to\infty}n\left(\sqrt[n]a-1\right)=\lim_{n\to\infty}\frac{a^\frac1n-1}{\frac1n},$$which is the derivative at $0$ of the function $t\mapsto a^t$. Since $a^t=\exp\bigl(t\log(a)\bigr)$, this derivative is $\log a$.

From $(1)$, you can deduce that$$\lim_{n\to\infty}2^n\left(\sqrt[2^n]a-1\right)=\log a\in(0,+\infty).$$But then$$\lim_{n\to\infty}\sqrt[n]{2^n\left(\sqrt[2^n]a-1\right)}=1,$$which means that$$\lim_{n\to\infty}\sqrt[n]{\sqrt[2^n]a-1}=\frac12.$$

José Carlos Santos
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  • Where can I find more about the first limit equation? – jnyan Nov 23 '17 at 10:11
  • @jnyan . see : https://math.stackexchange.com/questions/124667/how-to-find-lim-limits-n-to-infty-n-sqrtna-1?rq=1 – Almot1960 Nov 23 '17 at 10:13
  • @jnyan I've expanded my answer. – José Carlos Santos Nov 23 '17 at 10:27
  • Could you please comment a bit on: $\lim_{n\to\infty}\sqrt[n]{2^n\left(\sqrt[2^n]a-1\right)}=\lim_{n\to\infty}\sqrt[n]{\log a}$ ? It is not clear to me why it should be true that $\lim_{n\to \infty} \sqrt[n]{f(n)}=\lim_{n\to \infty} \sqrt[n]{\lim_{m\to\infty}f(m)}$. – Surb Nov 23 '17 at 20:20
  • @Surb I am just using the fact that if a sequence $(x_n){n\in\mathbb N}$ of elements of $(0,+\infty)$ is such that $\lim{n\to\infty}x_n=l>0$, then $\lim_{n\to\infty}\sqrt[n]{x_n}=1$. – José Carlos Santos Nov 23 '17 at 20:45
  • @JoséCarlosSantos Thanks, it makes now sense to me. But then I think that writing $\lim_{n\to\infty}\sqrt[n]{2^n\left(\sqrt[2^n]a-1\right)}=\lim_{n\to\infty}\sqrt[n]{\log a}$ is very misleading. The equality is true, but it is true because $\lim_{n\to\infty}\sqrt[n]{2^n\left(\sqrt[2^n]a-1\right)}=1$ and $1=\lim_{n\to\infty}\sqrt[n]{\log a}$. So using this equality to justify $\lim_{n\to\infty}\sqrt[n]{2^n\left(\sqrt[2^n]a-1\right)}=1$ doesn't feel correct. – Surb Nov 23 '17 at 21:26
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    @Surb I agree. I've edited my answer. Thanks. – José Carlos Santos Nov 23 '17 at 21:36
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Let's do this without logarithms, using only some algebra, the limit $\sqrt[n]x\to1$ for any $x\gt0$, and the Squeeze Theorem.

If $A\gt1$ and $N\in\mathbb{N}$, it's easy to see that

$$(A^N-1)=(A-1)(A^{N-1}+\cdots+1)\le(A-1)(A^N+\cdots+A^N)=NA^N(A-1)$$

and

$$A^N-1=(1+(A-1))^N-1=1+N(A-1)+\cdots+(A-1)^N-1\ge N(A-1)$$

hence

$${A^N-1\over NA^N}\le A-1\le{(A^N-1)\over N}$$

Now let $N=2^n$ and $A=a^{1/N}=a^{1/2^n}$ with $a\gt1$. Then

$${a-1\over2^na}\le a^{1/2^n}-1\le{a-1\over2^n}$$

and so

$${1\over2}\sqrt[n]{a-1\over a}\le\sqrt[n]{a^{1/2^n}-1}\le{1\over2}\sqrt[n]{a-1}$$

Since $a-1$ and $(a-1)/a$ are positive, the upper and lower bounds tend to $1/2$ as $n\to\infty$, so the Squeeze Theorem tells us

$$\lim_{n\to\infty}\sqrt[n]{a^{1/2^n}-1}={1\over2}$$

Barry Cipra
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The limit in question is reduced to the limit $$ \lim_{t \to 0_+} (a^t-1)^{1/\ln(t)}=e $$ which is worth remembering.

Olod
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