Show that $\lim _{x\rightarrow 1}\frac{\sum _{n=0}^{\infty }a_nx^{n\ }}{\frac{1}{1-x}\log \left(\frac{1}{1-x}\right)} = 1$ The problem is stated as:

Given that $(a_n)_{n=1}^{\infty}$ is a sequence for which $a_n/\log(n) \rightarrow 1$ as $n\rightarrow \infty$, show that $$\lim _{x\rightarrow 1}\frac{\sum _{n=0}^{\infty }a_nx^{n\ }}{\frac{1}{1-x}\log \left(\frac{1}{1-x}\right)} = 1$$


My attempt
First of all, we make the following substitution $\frac{1}{1-x} := u$, for which we get that $x = 1 - 1/u$. Hence, we can rewrite our limit as:
$$\lim _{u\rightarrow \infty}\frac{\sum _{n=0}^{\infty }a_n(1-1/u)^{n\ }}{u\log \left(u\right)}$$
Now, after this step, I'm really stuck. I've tried going for some Taylor expansion for $(1-1/u)^{n\ }$, but it only seems to make things worse. Furthermore, I tried to use the fact that $a_n/\log(n) \rightarrow 1$, so that for large n's, we have that our original sum is approximately:
$$\lim _{u\rightarrow \infty}\frac{\sum _{n=0}^{\infty }\log(n)(1-1/u)^{n\ }}{u\log \left(u\right)}$$
Since we have to show that the limit is $1$, we also have to somehow show that ${\sum _{n=0}^{\infty }\log(n)(1-1/u)^{n\ }} \approx u\log(u)$, however, I don't really know how to get there.
As a side note: it has also been given that $(\log n)^{-1}(1+1/2+...+1/n) \rightarrow 1$.

I'd appreciate if you could give me some hints that could help me solve this problem. Thank you.
 A: I'm going to use the notation @Gary used, i.e.
$$
\frac{1}{1-x}\log \left(\frac{1}{1-x}\right) =\sum_{n=1}^\infty H_nx^n,
$$
where $H_n = 1+\frac{1}{2}+\cdots+\frac{1}{n}$. As @Gary stated, note that $0<H_n-\log n\leq 1$ holds. These are not so difficult to prove, so you should try doing so.
Combined with $\lim_{n\to \infty}\frac{a_n}{\log n} =1$, we get $\lim_{n\to \infty}\frac{a_n}{H_n} =1$ . So, for any $\varepsilon >0$, there exists $N$ so that $n\geq N$ implies $|\frac{a_n}{H_n}-1|< \varepsilon$.
Fix $N$ and let $f(x) = \sum_{n=0}^{N-1}a_nx^n-\sum_{n=1}^{N-1}H_nx^n$ for simplicity. Since $N$ is a fixed finite number, note that $f(x)$ is bounded near $x=1$.
\begin{align}
\left | \frac{\sum_{n=0}^{\infty}a_nx^n}{\frac{1}{1-x}\log (\frac{1}{1-x})} -1 \right| &=  \left|\frac{\sum_{n=0}^\infty a_nx^n-\sum_{n=1}^\infty H_nx^n}{\sum_{n=1}^\infty H_nx^n}\right|\\
&\leq \frac{|f(x)|}{\sum_{n=1}^\infty H_nx^n} +\frac{|\sum_{n=N}^\infty (a_n- H_n)x^n|}{\sum_{n=1}^\infty H_nx^n} && \text{(by triangular inequality)} \\
&\leq \frac{|f(x)|}{\sum_{n=1}^\infty H_nx^n} + \frac{\sum_{n=N}^\infty |a_n- H_n|x^n}{\sum_{n=1}^\infty H_nx^n} && \text{(by triangular inequatlity)} \\
& \leq \frac{|f(x)|}{\sum_{n=1}^\infty H_nx^n} + \varepsilon \frac{\sum_{n=N}^\infty H_nx^n}{\sum_{n=1}^\infty H_nx^n} && \text{(by $|a_n - H_n|<\varepsilon H_n$)}\\
&\leq \frac{|f(x)|}{\sum_{n=1}^\infty H_nx^n} + \varepsilon
\end{align}
Recalling that $f(x)$ is bounded near $x=1$, $x \to 1-0$ implies the RHS $\to \varepsilon$. Therefore we get
$$
\limsup_{x\to 1-0}\text{LHS} \leq \varepsilon
$$
Since $\varepsilon$ is arbitrary positive number, we conclude that
$$
\lim_{x \to 1-0} \text{LHS} = 0.
$$
