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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.

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    This is also called Raabe's test. Related: https://math.stackexchange.com/questions/631348/proof-of-raabes-test2018-01-13

2 Answers 2

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Suppose $\,10\,\,s.t.\,\,q:=p-\epsilon>1$ . 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)

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Consider a real number $1. By convergence, there exists some integer $n_0$ such that $\forall n\geq n_0,~q so for all $n\geq n_0$, $\frac{|a_{n+1}|}{|a_n|}<1-\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.

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    no problem Conan :)2012-12-03