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$$\sum\limits_{n=1}^\infty \log(1+a_n) \text{ converges absolutely} \Leftrightarrow \sum_{n=1}^\infty a_n \text{ converges absolutely}.$$

How to prove this,

Suppose $$\sum_{n=1}^\infty a_n \text{ converges absolutely}.$$ Let $u_{n}=a_{n}$ and $v_{n}=\log(1+a_n)$, then $$\lim_{n\to\infty} \frac{u_{n}}{v_{n}}=1>0 \implies\sum_{n=1}^\infty \log(1+ a_n) \text{ converges absolutely}.$$ How to prove the converse part?

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  • $\begingroup$ This is an useful inequality jstor.org/stable/3615890?seq=1#page_scan_tab_contents $\endgroup$ – rtybase Oct 10 '17 at 17:10
  • $\begingroup$ But, interestingly, I believe it is false if you omit "absolutely". $\endgroup$ – GEdgar Oct 10 '17 at 17:19
  • $\begingroup$ if $$\lim_{n\to\infty } \log(1+a_n) \not=0 \Longleftrightarrow\lim_{n\to\infty }a_n \not = 0$$ then both series diverge. Assume that $a_n \to 0$, $\lim_{n\to\infty }\frac{|\log(1+a_n)|}{|a_n|} =\left|\lim_{h\to 0}\frac{\log(1+h)}{h}\right| = 1$ There is $n_0$ such that for $n>n_0$ $\left|\frac{|\log(1+a_n)|}{|a_n|} -1\right|<1/2 \Longleftrightarrow \frac12 |a_n|< |\log(1+a_n)|<\frac32|a_n|$ $\endgroup$ – Guy Fsone Nov 27 '17 at 15:41
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Hint: From the definition of $\ln'(1),$ we have

$$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1.$$

Thus there is $a>0$ such that

$$\frac{1}{2}\le \left|\frac{\ln (1+u)}{u}\right| \le \frac{3}{2}$$

for $u\in (-a,a),u\ne0.$

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The limit comparison test says that if you have two sequences $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ such that $\lim_{n\to \infty}\frac{a_n}{b_n}=c$ with $0<c<\infty$. Then $\sum_{n=1}^\infty a_n<\infty$ if and only if $\sum_{n=1}^\infty b_n<\infty$.

So we have to prove that $\lim_{n\to \infty}\frac{\vert ln(1+x_n)\vert}{\vert x_n\vert}=c$ with $0<c<\infty$. To do this we observe that $\lim_{x\to 0}\frac{\vert ln(1+x)\vert}{\vert x\vert}=1$ by the L'Hopital Rule. As in any of the two cases ($\sum_{n=1}^\infty x_n$ converges absolutely or $\sum_{n=1}^\infty ln(1+x_n)$ converges absolutely) we will have that $\lim_{n\rightarrow \infty}x_n=0$, then $\lim_{n\to\infty}\frac{\vert ln(1+x_n)\vert}{\vert x_n\vert}=1$.

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  • $\begingroup$ we have to prove what ? $\endgroup$ – Guy Fsone Oct 10 '17 at 17:11
  • $\begingroup$ That's not correct, since we don't know that $x_n \to 0$. $\endgroup$ – Adayah Oct 10 '17 at 17:17
  • $\begingroup$ @Adayah sure we conclude that, since otherwise the sum $\sum_{n=1}^k\ln(1+x_n)$ cannot converge. This and the answer of zhw. should be enough for the converse statement! $\endgroup$ – user190080 Oct 11 '17 at 10:47
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[This is essentially from @Guy Fsone's answer. However, it is rewritten in a significantly different way.]

First of all, observe that (by continuity of the exponential function and the logarithm) $$\lim_{n\to\infty } \log(1+a_n) =0 \Leftrightarrow \lim_{n\to\infty }a_n = 0.$$

  • If $\lim_{n\to\infty}a_n=0$ is not true, (note carefully that this assumption is fundamentally different from "$\lim_{n\to\infty}a_n\neq 0$"), then both $\sum_{n=1}^\infty |\log(1+a_n)|$ and $\sum_{n=1}^\infty |a_n|$ diverge.

  • Assuming that $a_n \to 0$ as $n\to\infty$, we have $$\lim_{n\to\infty }\frac{|\log(1+a_n)|}{|a_n|} =\left|\lim_{h\to 0}\frac{\log(1+h)}{h}\right| = 1\tag{1}$$ where we use the fact that the derivative of $x\mapsto \log x$ at $x=1$ is $1$ and the continuity of the absolute value function. Without loss of generality, we assume here that $|a_n|>0$ for all $n$.

    It follows from (1) that there exists $n_0$ such that for $n>n_0$

    $$\left|\frac{|\log(1+a_n)|}{|a_n|} -1\right|<1/2 $$ which implies that $$ \frac12 |a_n|< |\log(1+a_n)|<\frac32|a_n|~~~\forall ~~~n>n_0 $$ and thus $$ \frac12 \sum_{n>n_0}|a_n|< \sum_{n>n_0}|\log(1+a_n)|<\frac32 \sum_{n>n_0}|a_n|.$$

    This estimate completes the proof.

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    $\begingroup$ @GuyFsone: let me address your question one by one. Please be patient. (1)"I would like to know what is new here." As it was pointed out by a comment in meta, the logical structure of your original answer is heavily muddled. Since you have been keeping asking a hard evidence for what is wrong for your answer, let me explain in detail (in the next comment box). Again, please be patient before I finish my typing. $\endgroup$ – Jack Oct 22 '17 at 15:03
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    $\begingroup$ @GuyFsone: Let $A$ be the statement that $\sum_{n=1}^\infty\log(1+a_n)$ and $\sum_{n=1}^\infty a_n$ converge absolutely. Let $B$ be $\lim_{n\to\infty}\log(1+a_n)=0$ and let $C$ be $\lim_{n\to\infty} a_n=0$. The first sentence of your original answer says: $A\Rightarrow (B\Leftrightarrow C)$. Note that this a true statement, but does not contribute anything to your whole argument since $(B\Leftrightarrow C)$ is true regardless $A$ is true or not. (cont.) $\endgroup$ – Jack Oct 22 '17 at 15:14
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    $\begingroup$ @GuyFsone: (cont.) What you really want to say is $A\Rightarrow (B\hbox{ and } C)$. This is not a word playing game. This is a serious logic issue: "$A\Rightarrow (B\Leftrightarrow C)$" and "$A\Rightarrow (B\text{ and } C)$" are fundamentally different. (I'm still typing. Be patient.) $\endgroup$ – Jack Oct 22 '17 at 15:14
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    $\begingroup$ @GuyFsone I have already explained in detail what is wrong with your writing in your answer, which has serious logic problems. Your several comments show that you don't understand what I'm talking about at all and it seems that you don't want to. Without a commitment to mathematical logic, there is no way going on a meaningful discussion. If you want to insist that you are writing a correct thing and keep yelling how wrong others are, it is certainly your right. I will stop here. $\endgroup$ – Jack Oct 22 '17 at 15:33
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    $\begingroup$ @Jack: While most of what you say is correct, I have an alternative explanation for what is happening: Guy Fsone is not a native speaker of English, and this is painfully seen in his awkward and sometimes seemingly illogical formulations. What I'm saying is that the same answers formulated in his own native language would probably be correct. $\endgroup$ – Alex M. Oct 23 '17 at 9:01
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My way: $$\sum_{n=1}^\infty \log(a_n+1) = \log\prod_{n=1}^\infty (a_n+1)$$ since the series converges absolutely so does the product. Hence $a_n +1 \rightarrow 1$ and so $a_n \rightarrow 0$. Now the nontrivial step: $a_n \rightarrow 0$, and in proximity of $0$ you have $\log(1+x) = x + o(x^2)$ so dividing by $x$ for any $\epsilon > 0$ exists $n$ such that $$\frac{|\log(a_n+1)|}{|a_n|} \leq 1+ \epsilon$$ or in other words $$(1-\epsilon)|a_n| \leq |\log(a_n+1)| \leq (1+\epsilon)|a_n|$$ Apply the sum to the last inequality and you'll prove the proposition both ways.

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