Convergence of Dirichlet g.f. and the order of generating function

Question

Question Let $\sum_{n=1}^\infty\frac{a_n}{n^s}$ converge for a fixed positive $s$. Prove that $$\sum_{n=1}^\infty a_nx^n=o((1-x)^{-s}),\ x\to1^-.$$

Thoughts
I can prove this equation as long as $a_n=o(n^{s-1})$ (although it is false in general). Choose an $N( \varepsilon )$ s.t. forall $n>N( \varepsilon )$, $a_nn^{-s} |<\varepsilon$ or $| a_n |<\varepsilon n^s$. $$\sum_{n=1}^\infty{a_nx^n}=O( 1 ) +\sum_{n>N\left( \varepsilon \right)}{a_nx^n}\le O( 1 ) +\varepsilon \sum_{n>N( \varepsilon)}{n^{s-1}x^n}\\=O(1)+\varepsilon\operatorname{Li}_{1-s}(x)=O(1)+\varepsilon C(s)(1-x)^s$$ Therefore, I can prove it when $a_n$ is a non-negative sequence. Without assuming this, I can only prove the generating function of $a_n$ is $o((1-x)^{-s-1})$.

Answer 1

This problem is from the first chapter of a Chinese book. I don't know how to translate the title exactly, so I briefly call it Basis of Rank Estimation(written by XiuYuan Yu and ChengDong Pan).

Now we use the following lemma to complete the proof.

Lemma. Assuming that $b_n\to 0$ monotonically, and the series $\sum_{n=1}^{\infty}a_n b_n$ converges, then $$\lim_{n\to \infty}(a_1+a_2+\cdots+a_n)b_n=0.$$

Letting $b_n=\frac{1}{n^s}$ can get $$\lim_{n\to \infty}\frac{a_1+a_2+\cdots+a_n}{n^s}=0.$$ Put $\left(1-x\right)^{-s}=:\sum_{n=0}^{\infty}c_n x^n$, and note that \begin{align*} \left(1-x\right)^{-s-1}& =\frac{1}{1-x}\sum_{n=0}^{\infty}c_n x^n\\ & =\sum_{n=0}^{\infty}(c_1+c_2+\cdots+c_n)x^n. \end{align*} Thus $$\sum_{j=0}^n c_j=\binom{n+s}{s}\sim \frac{n^s}{s\Gamma(s)}, n\to \infty.$$ It follows that $$\sum_{j=1}^n a_j=o\left(n^s\right)=o\left(\sum_{j=0}^n c_j\right).$$ Hence as $x\to 1^-$, $$\sum_{n=1}^{\infty}a_n x^n=o\left(\sum_{n=0}^{\infty}c_n x^n\right)=o\left(\frac{1}{\left(1-x\right)^s}\right).$$