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$$ \sum_{n=1}^{\infty}|a_n| = C_1; \:|a_n| < 1 \implies \sum_{n=1}^{\infty}|ln(1+a_n)| = C_2 $$

i.e. second series is absolutely convergent if and only if first series is absolutely convergent.

My attempt:

(1) $$ |ln(1+a_n)| = |a_n + O(a_n^{2})| = |a_n| + O(a_n^{2}) $$ for |$a_n|<1$

So asymptotically

$$ \sum_{n=1}^{\infty}ln(1+a_n) = \sum_{n=1}^{\infty}|a_n| + O(\sum_{n=1}^{\infty}a_n^{2})$$ which should be convergent if (2) $ \sum_{n=1}^{\infty}|a_n|$ as $a_n^{2}$ is smaller than $|a_n|$.

Not sure about validness of step (1) and conclusion (2).

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  • $\begingroup$ Try the ratio test, i.e. $|\frac{\ln(1+a_n)}{a_n}|$ $\endgroup$ – polfosol Oct 15 '16 at 12:13
  • $\begingroup$ I thought about this, but this exact term doesn't seem to be correct (as far as I see it). But correct $\frac{|ln(1+a_n)|}{|a_n|}$ may quite large interval far bigger than 1 $\endgroup$ – Joe Half Face Oct 15 '16 at 12:19
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from l'hospital rule we get :$$lim_{x \to 0} \dfrac{ln(1+x)}{x} = 1 $$

now, we have $lim (a_n) = 0 $ so : $$lim_{n \to \infty} \dfrac{|ln(1+a_n)|}{|a_n|} =lim_{n \to \infty} |\dfrac{ln(1+a_n)}{a_n} | = |lim_{n \to \infty} \dfrac{ln(1+a_n)}{a_n}| = 1 $$

Note that I used the fact that the abs. function is continuous.

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  • $\begingroup$ Using L'Hopital is circular. That limit is just $\ln'(1).$ $\endgroup$ – zhw. Oct 15 '16 at 15:28

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