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Suppose that $\sum a_{n}$ is convergent series of real numbers. Either prove that $\sum b_{n}$ converges, or give a counter-example, when we define $b_{n}$ by $\frac{a_{n}}{1+|a_{n}|}$.

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  • $\begingroup$ As long as I didnt make a mistake, it is possible to prove with the ratio test. If I have time later, I will add that. $\endgroup$ – CBenni Apr 2 '13 at 9:44
  • $\begingroup$ @CBenni it is not possbile with the ratio test, as that would imply absolute convergence $\endgroup$ – Dominic Michaelis Apr 2 '13 at 9:47
  • $\begingroup$ @DominicMichaelis then I wont even bother ;) I mustve done a minor mistake somewhere. $\endgroup$ – CBenni Apr 2 '13 at 19:13
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For series that have some negative terms, this is not necessarily true. We give a family of counterexamples.

Consider the series $$a_1+a_1-2a_1+a_2+a_2-2a_2+a_3+a_3-2a_3+\cdots,$$ where the $a_i$ are a sequence of positive terms slowly decaying to $0$. The convergence of the above series is clear.

The modified series need not converge.

For the modified series looks like $$\frac{a_1}{1+a_1}+\frac{a_1}{1+a_1}-\frac{2a_1}{1+2a_1}+\frac{a_2}{1+a_2}+\frac{a_2}{1+a_2}-\frac{2a_2}{1+2a_2}+\cdots.$$ If we add terms by threes, we get $$\sum_{i=1}^\infty \frac{2a_i^2}{(1+a_i)(1+2a_i)}.$$ Choose for instance $a_i=\dfrac{1}{i^{1/8}}$, and we get divergence.

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  • $\begingroup$ So the answer is that $\sum\frac{a_{n}}{1+|a_{n}|}$ doesn't have to converge. Thanks. $\endgroup$ – Crni Apr 2 '13 at 11:02
  • $\begingroup$ You are welcome. Yes, that's the answer. Easily we have convergence if the terms are all $\ge 0$. $\endgroup$ – André Nicolas Apr 2 '13 at 11:07
  • $\begingroup$ @AndréNicolas ah great I knew I could trust my intuition but still my lazyness was to strong to search for a counterexample $\endgroup$ – Dominic Michaelis Apr 2 '13 at 19:17
  • $\begingroup$ and not to forget +1 for the excellent answer $\endgroup$ – Dominic Michaelis Apr 2 '13 at 19:26

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