$\sum_{n=1}^{\infty}|ln(1+a_n)|$ is convergent only when $\sum_{n=1}^{\infty}|a_n|$ is convergent for $|a_n| < 1$

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

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).

• Try the ratio test, i.e. $|\frac{\ln(1+a_n)}{a_n}|$ – polfosol Oct 15 '16 at 12:13
• 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 – Joe Half Face Oct 15 '16 at 12:19

from l'hospital rule we get :$$lim_{x \to 0} \dfrac{ln(1+x)}{x} = 1$$
now, we have $lim (a_n) = 0$ so : $$lim_{n \to \infty} \dfrac{|ln(1+a_n)|}{|a_n|} =lim_{n \to \infty} |\dfrac{ln(1+a_n)}{a_n} | = |lim_{n \to \infty} \dfrac{ln(1+a_n)}{a_n}| = 1$$
• Using L'Hopital is circular. That limit is just $\ln'(1).$ – zhw. Oct 15 '16 at 15:28