Proof $\lim\limits_{n \rightarrow \infty} \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0$

Question

How would one go about proving the limit of sequence: $$\lim_{n \rightarrow \infty}\ \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0\,. $$ Epsilon definiton of limit seems to be too complicated and/or unsolvable (variable both in exponent and base). $\lim\limits_{x \rightarrow \infty} \ 3^{\frac{1}{n}}$ is clearly $1$, but how does it combine with $\sqrt{n}$?

Is there an obvious upper bound that I'm missing?

Answer 1

From $$1=3-2=\left(3^{\frac{1}{n}}\right)^n-\left(2^{\frac{1}{n}}\right)^n=\\ \left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)\left( 3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}\right)$$ we have $$0<\sqrt{n}\left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)=\frac{\sqrt{n}}{3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}<\\ \frac{\sqrt{n}}{2^{\frac{n-1}{n}}+2^{\frac{n-2}{n}} 2^{\frac{1}{n}}+2^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+2^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}= \frac{\sqrt{n}}{n2^{\frac{n-1}{n}}}< \frac{1}{\sqrt{n}}$$ Or $$0<\sqrt{n} \left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)<\frac{1}{\sqrt{n}}$$ And use the squeeze theorem.

Answer 2

Simply use the definition of $a^x$ and Taylor's formula at order $1$: \begin{align} \sqrt n\Bigl(3^{\tfrac1n}-^{\tfrac1n}\Bigr)&=\sqrt n\Bigl(\mathrm e^{\tfrac{\log 3}n}-\mathrm e^{\tfrac{\log2}n}\Bigr)=\sqrt n\biggl(1+\frac{\log 3}n+o\Bigl(\frac1n\Bigr)-1-\frac{\log 2}n-o\Bigl(\frac1n\Bigr)\biggr)\\ &=\frac{\log 3-\log2}{\sqrt n}+o\Bigl(\frac1{\sqrt n}\Bigr)\to 0. \end{align}

Answer 3

$$\lim_{z\to 0}\frac{3^z-2^z}{z}=\log\frac{3}{2} $$ can be proved in a number of ways, for instance through De l'Hopital rule. It leads to $$ 3^{1/n}-2^{1/n} \leq \frac{C}{n} \qquad (C>0)$$ as $n\to +\infty$, hence the given limit is zero by squeezing.

Answer 4

One possibility is evaluate $$\lim_{x\to 0}\dfrac{3^x-2^x}{\sqrt{x}}$$ using L'Hospital's Rule rule. Or otherwise we can use the identity $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + … + x y^{n-2} + y^{n-1})$ to obtain $$\sqrt{n} (3^{1/n} - 2^{1/n})=\dfrac{\sqrt{n}}{3^{(n-1)/n} + 3^{(n-2)/n} 2^{1/n} + … + 3^{1/n} 2^{(n-2)/n} + 2^{(n-1)/n}}.$$ Then $$\dfrac{3^{(n-1)/n} + 3^{(n-2)/n} 2^{1/n} + … + 3^{1/n} 2^{(n-2)/n} + 2^{(n-1)/n}}{n}\gt 2^{(n-1)/n}$$ as average is larger than the smallest term. Now use squeeze theorem.

Answer 5

Consider $f(x) = x^{(1/n)}.$

$\dfrac{f(3) - f(2)}{1} = f'(t)$, $2 \lt t\lt3$

$ f'(t) = \dfrac{1}{nt^{(1-1/n)}}.$

Let $n \ge 2:$

$a_n:= √n(3^{(1/n)} -2^{(1/n)}) \lt \dfrac{√n}{n2^{(1/2)}}$

$= \dfrac{√n}{n} =\dfrac{1}{√n}.$

Let $\epsilon >0$ be given.

There is a $n_0 \gt 1/\epsilon^2$. (Archimedes)

For $n\ge n_0 :$

$|a_n| \lt \dfrac{1}{√n} \le \dfrac{1}{√n_0} \lt \epsilon.$