I have been having fun with infinite power towers and I came across this problem:
Let $a_2=\frac12$, and for $n>2$, $a_n=a_{n-1}^{\frac{1}{n}}$. Find $\lim_{n\to\infty}a_n$.
So I have absolutely no idea how to deal with the fractional exponents. I tried the standard power tower method where you assign it to a variable but I can't get anywhere with that. Does anyone have any ideas?
You have look at the first terms of the recurrence sequence to be able to understand the explicit expression of $a_n$:
$a_3=\frac{1}{2}^{\frac{1}{3}}=\frac{1}{4}^{\frac{1}{6}}$
$a_4=({\frac{1}{2}^{\frac{1}{3}}})^{\frac{1}{4}}=\frac{1}{4}^{\frac{1}{24}}$
$a_5=\{[({{\frac{1}{2}^{\frac{1}{3}}})]^{\frac{1}{4}}}\}^{\frac{1}{5}}=\frac{1}{4}^{\frac{1}{120}}$
. . .
$a_n=\frac{1}{4}^{\frac{1}{n!}}$.
So $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{4}^{\frac{1}{n!}}=1$ (since $\lim_{n\to\infty}\frac{1}{n!}=0$).