I recently came across this question here and there is something I don't understand.
In the question we are considering a sequence $a_n$ with $\lim \sup_{n \to \infty}a_n \le \rho$ and we want to show that $\lim \sup_{n \to \infty} a_n^{{(n-m)}/{n}} \le \rho$. The asker tried to split up the $a_n^{{(n-m)}/{n}}$ as
$$ a_n^{{(n-m)}/{n}} = a_n \cdot a_n^{- {m \over n}}$$
but the answerer gives an example for which this does not work: if $a_n = \frac{1}{n^{\frac{n}{m}}}$ then $a_n^{- {m \over n}}$ diverges.
What I don't understand is why this example works. Because for sequences that converge (and $\limsup$ of a sequence always converges) we have
$$ \lim a_n b_n = \lim a_n \lim b_n$$
so it seems to me that splitting up a product should always be allowed. Yet, in this case it's not.
Please could someone explain to me what I'm missing and why the product rule for limits cannot be applied here?
With $a_n=1/(n^{n/m}$ and $b_n=a_n^{-m/n}=n$ we have $\lim a_n=0$ The expression $\lim b_n=\infty$ is symbolic short-hand and so is the sentence "$b_n$ goes to $\infty$" because there is no number $\infty$ for which all the usual rules of arithmetic apply. In particular "$\lim a_n .\lim b_n=0.\infty$" is meaningless. The product-limit rule is only valid when the limits are both real numbers.