What is $\lim_{n\to\infty} n(2^{1/n} -1)$

How would you find the value of $\lim_{n\to\infty} n(2^{1/n} -1)$?

I tried taking the logarithm on both sides and it does not appear to be working.

• Hint What's the derivative of $2^x$ at $x=0$?
– Pedro
Commented Mar 23, 2015 at 6:30
• We have in general for $a>0$ that $$\lim_{x\rightarrow 0}\frac{a^x-1}{x}=\ln a$$ This is a useful limit to keep in hand.
– user123641
Commented Mar 23, 2015 at 20:33

You can go this way.

$$n(2^{1/n}-1 ) = n( e^{\frac{\ln(2)}{n}} - 1 ) = n \left( \left(1+\frac{\ln(2)}{n}+\frac{\ln^2(2)}{2!\,n^2}+\dots \right)-1 \right)$$

$$= {\ln(2)}+\frac{\ln^2(2)}{2!\,n}+\dots \longrightarrow_{n\to \infty} \ln(2) .$$

Note: You can use the big O notation if you are familiar with it

$$e^{\frac{\ln(2)}{n}} = 1+\frac{\ln(2)}{n}+O\left( \frac{1}{n^2}\right)$$

• How do you know that limit apply termwise?
– user170039
Commented Mar 23, 2015 at 6:54
• @user170039, uniform convergence of the Taylor series in disks of radius < radius of convergence. Commented Mar 23, 2015 at 9:54
• Typo: missing factorials in the denominators of the series. Commented Mar 23, 2015 at 9:58
• @Martín-BlasPérezPinilla: Thanks for providing an answer for his comment. Sometimes I do not give such details because the OP's do not have a sufficient background or it is advanced stuff for them You can see it is a limit problem and I do not think the OP have a background about uniform convergence since these things ae taught at advanced level. Commented Mar 23, 2015 at 20:15

Write $f(x)=2^x=e^{x\log{2}}$ then $f'(x)=2^x\log{2}$. We have

$$n(2^{\frac{1}{n}}-1)=\frac{f(a_n)-1}{a_n}$$

Avec $a_n=\frac{1}{n}\rightarrow 0$. So

$$\lim_{n\to \infty}n(2^{\frac{1}{n}}-1)=f'(0)=\log{2}$$

Set $x = \dfrac{1}{n}$ , so $\lim_{n \to \infty} x = 0$

Therefore =

$$\lim_{n\to \infty} n\left(2^{\frac{1}{n}}-1\right) = \lim_{x\to 0} \dfrac{\left(2^{x}-1\right)}{x} = \lim_{x\to 0} \ln 2 \dfrac{\left(e^{x\ln2}-1\right)}{x\ln2}$$

Now we do $m = x\ln2$ so $lim_{x \to 0} m = 0 \therefore$

$$\lim_{x\to 0} \ln 2 \dfrac{\left(e^{x\ln2}-1\right)}{x\ln2} = \ln 2\lim_{m\to 0} \dfrac{\left(e^{m}-1\right)}{m}$$

The limit is a classic and it's value is 1 (http://www.ies-math.com/math/java/calc/lim_e/lim_e.html), so

$$\lim_{n\to \infty} n\left(2^{\frac{1}{n}}-1\right) = \ln 2$$

Let $x>1$. From $$\log(x)=\int_1^x\frac{dt}{t}$$ it follows that $$1-\frac{1}{x}<\log(x)<x-1$$ and $$\log(x)=n\log(x^{1/n}).$$ Then $$n(1-x^{-1/n})<\log(x)<n(x^{1/n}-1)$$ and so $$x^{-1/n} < \frac{\log(x)}{n(x^{1/n}-1)} < 1.$$ Now let $n\to\infty$.