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.


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


and since


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)


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.

  • $\begingroup$ no problem Conan :) $\endgroup$ – Olivier Bégassat Dec 3 '12 at 18:18

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