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.
Best Answer
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)