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

Solution 1:

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 $$ \lim_{x \to 1-0} \text{LHS} = 0. $$