Find limit of $a_n = \sqrt[n!]{\frac{1}{2^{(n!)}}-\frac{1}{3^{(n!)}}}$

Question

Given $a_n = \sqrt[n!]{\frac{1}{2^{(n!)}}-\frac{1}{3^{(n!)}}}$. I have to find its limit. Of course $a_n \le \sqrt[n!]{\frac{1}{2^{(n!)}}} = \frac{1}{2}$ but it's not enough to use squeeze theorem. Any hints?

Answer 1

As $n \to \infty$ we have $n! \to \infty$ so this is analogous to asking: $\lim_{x \to \infty} \left( \frac{1}{2^x} - \frac{1}{3^x}\right)^{\frac{1}{x}}$ $$\lim_{x \to \infty} \left( \frac{1}{2^x} - \frac{1}{3^x}\right)^{\frac{1}{x}} = \lim_{x \to \infty} \frac{1}{2}\left( 1 - (\frac{2}{3})^x\right)^{\frac{1}{x}}$$

As $x \to \infty$ then $(2/3)^x \to 0$ and $(1/x) \to 0$ so that limit becomes: $$\lim_{x \to \infty} \frac{1}{2}\left( 1 - (\frac{2}{3})^x\right)^{\frac{1}{x}} = \frac{1}{2}(1-0)^0 = \frac{1}{2}$$

Answer 2

Note that $n!$ is a subsequence of the natural numbers. Hence, if we can find $\lim_{n \to \infty} \sqrt[n]{2^{-n} - 3^{-n}}$, then for any subsequence of the natural numbers $n_k$, the limit $\lim_{k \to \infty} \sqrt[n_k]{2^{-n_k} - 3^{-n_k}}$ is exactly the same.

Note that $\left(2^{-n} - 3^{-n}\right)^{\frac 1n} = \frac{(3^n - 2^n)^{\frac 1n}}{6}$, after cross multiplying and simplifying.

Now, note that $(3^n - 2^n)^{\frac 1n} = 3\left( 1 - \left(\frac 23\right)^n\right)^{\frac 1n}$. The limit of $(1 - x^n)^{\frac 1n}$ as $n \to \infty$ is $1$ for $|x| < 1$, which is true in our case. Hence, we see that $(3^n - 2^n)^{\frac 1n} \to 3$ as $n \to \infty$. Hence, we have that the desired limit is $\frac 36 = 0.5$.

Therefore, the limit $\lim_{n \to \infty} a_n = 0.5$.