Proving absolute convergence of a series assuming that the limit $\lim\limits_{k \to \infty} k\left(1-\frac{|a_{k+1}|}{|a_k|}\right)$ exists

Question

Let $\{a_k\}$ be a sequence of non-zero real numbers and suppose that

$$p = \lim_{k \to \infty} k\left(1-\frac{|a_{k+1}|}{|a_k|}\right)\quad\text{exists}$$

Prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely when $p > 1$.

I've tried manipulating the equation to isolate $\frac{|a_{k+1}|}{|a_k|}$ and use the Ratio Test. But it doesn't seem to work because then $\lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} = 1$, and the Ratio Test is inconclusive.

Probably some other Test (like the Logarithmic Test?) needs to be used but I'm unsure how.

Any advice would be appreciated. Thanks.

Answer 1

Suppose $\,1<p\,\Longrightarrow \exists\,\epsilon>0\,\,s.t.\,\,q:=p-\epsilon>1$ . For all but a finite number of indexes $\,k\,$ we have

$$k\left(1-\frac{|a_{k+1}|}{|a_k|}\right)>q\Longrightarrow \frac{|a_{k+1}|}{|a_k|}<1-\frac{q}{k}\leq \left(1-\frac{1}{k}\right)^q $$

where the last inequality follows from Bernoulli's Inequality

But then

$$\frac{|a_{k+1}|}{|a_k|}\leq\frac{(k-1)^q}{k^q}=\frac{\frac{1}{k^q}}{\frac{1}{(k-1)^q}}=\frac{b_{k+1}}{b_k}$$

and since

$$\sum_{k=1}^\infty\frac{1}{k^q}$$

converges the so does our series by the second comparison test (theorem 6.1, page 60, in the book "Sequences and series" in this place)

Answer 2

Consider a real number $1<q<p$. By convergence, there exists some integer $n_0$ such that $$\forall n\geq n_0,~q<n\bigg(1-\frac{|a_{n+1}|}{|a_n|}\bigg)$$ so for all $n\geq n_0$, $\frac{|a_{n+1}|}{|a_n|}<1-\frac{q}{n}<e^{-\frac{q}{n}}$. Multiplying these together, we get, for all $n\geq n_0$, $$|a_{n+1}|<|a_{n_0}|e^{-q\sum_{k=n_0}^{n}\frac{1}{k}}.$$ Now $\sum_{k=n_0}^{n}\frac{1}{k}=\ln(n)+O(1)$, so $$e^{-q\sum_{k=n_0}^{n}\frac{1}{k}}=\frac{O(1)}{n^q}$$ which is the term of a convergent series (because $q>1$) so $\sum |a_n|$ converges.