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 $a>1$ 
$$\underset{n\rightarrow\infty}\lim{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}=?, \;\;a>1$$

In Shaum's Mathematical handbook of formulas and tables I've seen:
$$\;\;\;\;\;\;\;\;\;\;\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\;,x\in\langle-1,1]\;\;\;\;\;\;\;$$
$$\frac{1}{2}\ln{\Bigg(\frac{1+x}{1-x}\Bigg)}=1+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots\;\;\;,x\in\langle-1,1\rangle$$
The term in parentheses reminded me of the harmonic series. I thought of using the Taylor series. Is that a good idea?
It says $a>0$ so I probably can't use these two formulas.
On the other hand:
$$e^x=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\;\;\;\;\;\;,$$
but there are no factorials in the denominators.
Source in Croatian: 2.kolokvij, matematička analiza

$$
\begin{align}
\lim_{n\to\infty}\frac{n}{a^{n+1}}\left(a+\frac{a^2}2+\cdots+\frac{a^n}n\right)
&=\lim_{n\to\infty}\left(\frac1a+\frac{n}{n-1}\frac1{a^2}+\frac{n}{n-2}\frac1{a^3}+\cdots\right)\tag1\\
&=\frac1a+\frac1{a^2}+\frac1{a^3}+\cdots\tag2\\
&=\frac1{a-1}\tag3
\end{align}
$$
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 $(2)$.
