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

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