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I am currently taking a course in integral calculus and I came across the so-called logarithmic criterion to study the convergence of a series. It is stated as follows. Let $a_n$ be a stricly positive sequence of real numbers. Let $$L = \lim_{k \rightarrow\infty} \frac {-\ln(a_k)}{\ln(k)}$$. If $L > 1$, then $$\sum_{k=1}^\infty a_k$$ is convergent, and if $L < 1$, $$\sum_{k=1}^\infty a_k$$ is divergent. Do any of you know how to prove it? Any feedback will be appreciated.

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  • $\begingroup$ $\lim$... what? What is the limit to? $\infty$? $\endgroup$ – Simply Beautiful Art Sep 14 '16 at 22:16
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    $\begingroup$ You better divide by $k$ and not $\log k$. And compare to the value 0 rather than 1. Then it is the root criterion (and you don't need a limit a lim inf suffices) $\endgroup$ – H. H. Rugh Sep 14 '16 at 22:21
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One direction: $ L<1 => \exists n \in \mathbb{N}\frac{-ln(a_{k})}{ln(k)}<1 \quad\forall k>n$ thus for those k: $-ln(a_{k})<ln(k) \quad (w.l.o.g \quad k>1 $ and therefore the logarithm positive) now take the exponential of both sides: => $\frac{1}{a_{k}}<k => a_{k}>1/k $ so the series diverges by the direct comparision test.

For $L>1$ you can show the same way, that $a_{k}<1/k$ and with some modifications probably that $a_{k}<\frac{1}{nk}$ for any fixed factor n, i'm not quite sure wheter this implies convergence, though.

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  • $\begingroup$ For L > 1 I finally figured out a way to prove convergence with a very similar idea to the one you used for L < 1. Basically you show that for all positive epsilon, there is an n such that from that point on, a_k < 1/k^(L-eps). Taking a suficiently small epsilon, since L >1, you are comparing the sequence with a power series with exponent greater than one, so it must converge. $\endgroup$ – Mr Peanutbutter Sep 15 '16 at 7:11

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