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Given $\displaystyle \sum_{n=1}^\infty a_n$ is absolutely convergent. Show that $\displaystyle \sum_{n=1}^\infty \dfrac{a_n}{1+a_n}$ also converges absolutely. (If $a_n \neq -1, \forall n \geq 1$ )

I am not sure where to start with this. Any help would be appreciated

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marked as duplicate by Did sequences-and-series Apr 24 '16 at 18:09

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    $\begingroup$ You know $a_n\to0$. So there exists $N$ such that $a_n>-1/2$ for all $n>N$... $\endgroup$ – David C. Ullrich Apr 24 '16 at 17:54
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We have: $\dfrac{|a_n|}{|1+a_n|}\leq \dfrac{|a_n|}{1-|a_n|}\leq \dfrac{|a_n|}{1-\frac{1}{2}}= 2|a_n| $ for all $n \geq N_0$ since $a_n \to 0$, and using comparison test, the conclusion follows.

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  • $\begingroup$ I understand the above inequality, but how do we know 2|an| is convergent. If |an|<1/2 then 2|an| is < 1. How does this prove convergence? $\endgroup$ – Harry Apr 24 '16 at 18:38
  • $\begingroup$ If $\displaystyle \sum_{k=1}^n |a_k| \to L \implies \displaystyle \sum_{k=1}^n 2|a_k| \to 2L$ $\endgroup$ – DeepSea Apr 24 '16 at 18:50
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From the fact that $\sum_n |a_n|$ converges, one has that $\lim_{n \rightarrow \infty} a_n = 0$. Then you can use the limit comparison test, comparing ${\displaystyle \sum_n \bigg|{a_n \over 1 + a_n}\bigg|}$ to $\sum_n |a_n|$, to show that your series is absolutely convergent, and therefore convergent.

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  • $\begingroup$ Using the limit comparison test, I would get lim n→∞[ an/(1+an)] / an. Which is equal to lim 1/(1+an). Since an -> 0 then the limit of 1/(1+an) =1 so the series an/(1+an) converges. Is this correct? $\endgroup$ – Harry Apr 24 '16 at 18:10
  • $\begingroup$ It's right, except you should put everything in absolute values, since the limit comparison test applies only to series with nonnegative terms. $\endgroup$ – Zarrax Apr 24 '16 at 18:45

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