Absolute Convergence of a Series (maybe use ratio test)

Question

Let

$p:=\lim_{{k \to \infty}} k(1-|\frac{a_{k+1}}{a_k}|)$

Prove

$p>1$ or $p=\infty \implies \sum_{n=1}^\infty a_n $ converge absolutely

Answer 1

Lemma 1: If $s\notin\mathbb{Z}_{\le 0}$, then $$\lim\limits_{x\to 0}\frac{(1+ax)(1-x)^a-1}{x^2} $$ exists and is non-zero.

proof: $(1+ax)(1-x)^a=\left(1+ax\right)\left( 1-ax+\frac{a(a-1)}{2}x^2+o(x^2)\right) =1-\frac{a^2+a}{2}x^2+o(x^2)$ which is the rephrasement of our lemma.

Lemma 2: If $s\notin\mathbb{Z}_{\le 0}$, then $$\lim\limits_{n\to\infty}\frac{s(s+1)\cdots(s+n)}{n^s\cdot n!} $$ exists and is not zero.

proof: Let $$b_n=\frac{s(s+1)\cdots(s+n)}{n^s\cdot n!}.$$ Then\begin{align*} \frac{b_{n+1}}{b_n}=&\left( 1+\frac{s}{n+1}\right) \left( 1-\frac{1}{n+1}\right)^s\\=&1+\frac{c}{n^2}+o\left(\frac{1}{n^2} \right) \text{ for some constant $c$ .} \end{align*} (The last step uses lemma 1 and sequence being $o\left(\frac{1}{(n+1)^2}\right)$ are also $o\left( \frac{1}{n^2}\right) $.)

Then $b_n=b_1\prod_{i=2}^n \frac{b_i}{b_{i-1}}=e^{\sum_{i=2}^n \ln\frac{b_{n_+1}}{b_n}}$. Since $\ln\frac{b_{n_+1}}{b_n}\sim \left( \frac{b_{n_+1}}{b_n}-1\right)\sim \frac{c}{n^2}$ we conclude $\sum_{i=2}^n \ln\frac{b_{n_+1}}{b_n}$ converges, which implies $b_n$ converges to a non-zero real number.

Finally we return to the question. The hypothesis implies that there exists $d>1$ s.t. for sufficiently large $k$, $\bigg|\frac{a_{k+1}}{a_k} \bigg|<1-\frac{d}{k}$ $\implies$ $$a_k<M\prod_{l=N}^k \left( 1-\frac{d}{k}\right)=M\prod_{l=N}^k\left(\frac{k-d}{k} \right) $$ for some $M$, $N$. Lemma 2 says $a_n\sim Mc\frac{n^{-d}n!}{n!}\sim A\cdot n^{-d}$ for some $A$. Since $d<1$ we conclude $\sum_{n=1}^{\infty} |a_n|<+\infty$.