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This question already has an answer here:

Prove $\sum_{n=1}^\infty a_n \in \mathbb R \mathbb \leftrightarrow \sum_{n=1}^\infty\frac{a_n}{1+a_n}\in \mathbb R$
The hint says that for one direction argue that $b_n = \frac{a_n}{\frac{a_n}{1+a_n}}$ is bounded after some fixed cutoff point. I dont know where the $b_n$ is coming from or which test to use to prove this.

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marked as duplicate by Martin R, Daniel Fischer real-analysis May 11 '17 at 13:05

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  • $\begingroup$ Note that the assertion need not hold if one doesn't assume $a_n \geqslant 0$. Expanding $\frac{1}{1+a_n}$, we find $$\frac{a_n}{1+a_n} = a_n - a_n^2 + \frac{a_n^3}{1+a_n}.$$ If $(a_n)$ is an alternating sequence such that $\sum a_n$ converges, $\sum a_n^2$ diverges, and $\sum a_n^3$ converges absolutely, then $\sum \frac{a_n}{1+a_n}$ diverges. $\endgroup$ – Daniel Fischer May 11 '17 at 13:12
  • $\begingroup$ @DanielFischer So why did you mark this as a duplicate, since $a_n \ge 0$ is not mentioned in the hypotheses? $\endgroup$ – zhw. May 11 '17 at 21:17
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    $\begingroup$ @zhw. It happened because I absentmindedly clicked in the wrong location. I hadn't yet made up my mind whether I should close it as a duplicate. Then the question arose whether to reopen. I didn't do that because I expect that the OP here just forgot to mention that assumption, though it was made in the exercise (but, unfortunately that isn't certain, there are enough wrong exercises posed). Ideally, the OP here would confirm whether the assumption was made in the exercise (and if not, it's a clear-cut reopening). $\endgroup$ – Daniel Fischer May 11 '17 at 21:34
  • $\begingroup$ @DanielFischer Excellent response. Thanks, MSE is lucky to have you. $\endgroup$ – zhw. May 11 '17 at 21:51
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suppose $\sum_{n=1}^\infty a_n/(1+a_n)$ converges then $lim_n a_n/(1+a_n)=0$ which implies $a_n\rightarrow 0$. Thus by limit comparison test $\sum_{n=1}^\infty a_n$ converges. For the other direction we directly get $a_n\rightarrow 0$ and hence again limit comparison test gives $\sum_{n=1}^\infty a_n/(1+a_n)$ converges .

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  • $\begingroup$ Are you assuming that $a_n > 0$? $\endgroup$ – Omnomnomnom May 11 '17 at 13:00
  • $\begingroup$ no i do not. i assume $a_n\neq -1\forall n$ $\endgroup$ – user379195 May 11 '17 at 13:03
  • $\begingroup$ Ah! I confused "limit comparison" for "direct comparison". Nicely done. $\endgroup$ – Omnomnomnom May 11 '17 at 13:05
  • $\begingroup$ Limit comparison requires an eventually fixed sign. For $c_n = \frac{(-1)^{n-1}}{\sqrt{n}}$ and $d_n = c_n + \frac{1}{n}$, we have $\frac{d_n}{c_n} \to 1$, but $\sum c_n$ is convergent while $\sum d_n$ is divergent. $\endgroup$ – Daniel Fischer May 11 '17 at 13:08
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First, note that $b_n = 1 + a_n$. Note moreover that $$ \sum_{n=1}^\infty a_n = \sum_{n = 1}^\infty \frac{a_n}{1 + a_n}b_n $$ The hint is trying to get you to use a comparison test, noting that $b_n$ is bounded.

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