How to prove $\lim_{n \to \infty} \frac{\log(n)(n^{\log(n)})} {a^n} ＝0, $ where $ |a|>1$ How to prove $$\lim_{n \to \infty} \frac{\log(n)\left(n^{\log(n)}\right)} {a^n} ＝0,\;  \text{ where } \;|a|&gt;1$$

I don't know how to employ L'Hospital here.
I don't know how to employ L'Hospital here.


Then, don't...
$$
\left|\frac{\log(n)\left(n^{\log(n)}\right)} {a^n}\right| ＝\exp\left(\log\log n+(\log n)^2-\alpha n\right)$$
with $\alpha=\log|a|\gt0$, and
$$
\log\log n+(\log n)^2-\alpha n\sim-\alpha n\to-\infty.
$$
Let $a_n:=\log(n)n^{\log(n)}$. 

First compute $\lim_n \frac{\log(n+1)}{\log(n)}=1$. We can use L'Hospital for this (It gets reduced to $\lim_n \frac{n}{n+1}=1$).

From this we get that $\lim_n [\log(n)]^{1/n}=1$ as well. This time by applying L'Hospital for sequences (Stoltz's theorem) which implies that if $\lim \frac{r_{n+1}}{r_n}=L$ then $\lim_n r_n^{1/n}=L$.

Now we compute $\lim_n n^{\log(n)/n}=\lim_{n}e^{\frac{\log^2(n)}{n}}=e^{\lim_n\frac{\log^2(n)}{n}}=1$. This we get by applying L'Hospital to the limit in the exponent $\lim_n\frac{\log^2(n)}{n}=0$, since by L'Hospital it gets reduced to $\lim_n\frac{2\log(n)}{n}$, which gets reduced to $\lim_{n}\frac{2}{n}=0$ by another application of L'Hospital.

Therefore, since $\lim_n[\log(n)]^{1/n}=1$ and $\lim_nn^{\log(n)/n}=1$, we get that $$\lim_n a_n^{1/n}=1.$$

Therefore, the radius of convergence of $\sum_n a_nx^n$ is at least $1$. Therefore $\sum_n \frac{a_n}{a^n}$ converges. Hence $$\lim_n\frac{a_n}{a^n}=0.$$