Finding $\lim_{n\to\infty}{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}$ where $a>1$

By Stoltz-Cesaro

$$\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+…+\frac{a^n}{n}\right)=\frac{\left(a+\frac{a^2}{2}+\frac{a^3}{3}+…+\frac{a^n}{n}\right)}{\frac{a^{n+1}}{n}}$$

we obtain

$$\frac{\frac{a^{n+1}}{n}}{\frac{a^{n+2}}{n+1}-\frac{a^{n+1}}{n}}=\frac1{\frac{na}{n+1}-1} \to \frac1{a-1}$$

$$ \begin{align} \lim_{n\to\infty}\frac{n}{a^{n+1}}\left(a+\frac{a^2}2+\cdots+\frac{a^n}n\right) &=\lim_{n\to\infty}\left(\frac1a+\frac{n}{n-1}\frac1{a^2}+\frac{n}{n-2}\frac1{a^3}+\cdots\right)\tag1\\ &=\frac1a+\frac1{a^2}+\frac1{a^3}+\cdots\tag2\\ &=\frac1{a-1}\tag3 \end{align} $$ The series on the right side of $(1)$ is dominated by $$ \frac1a+\frac2{a^2}+\frac3{a^3}+\cdots=\frac{a}{(a-1)^2}\tag4 $$ which validates $(2)$.