Does the transformation $a_n \leadsto \frac{a_n}{1+a_n}$ unconditionally preserve divergence of a series?

Question

If $\sum{a_{n}}$ is divergent, prove that $\sum\frac{a_{n}}{1+a_{n}}$ is divergent.

Is this always true? Because I found in the books this is just when $a_{n}>0$, and when I use $a_{n}=(r_{1},r_{2},-1,r_{4},-1,...)$, can we still say that the second series is divergent? Where $r_{i}$ can be any real such that $(a_{n})$ converges to zero. (The position of the $-1$ doesn't matter, it is just there to make the second series not defined).

But $a_{n} = (1)$ isn't divergent, is it? - so you haven't found a counterexample. — Clarinetist, Jun 25 '14 at 03:45
Sorry it was -1. So the sequence is not even defined, that was my question if we can consider that case as divergence. — Jorge Ramos, Jun 25 '14 at 03:46
you are right, my mistake but consider a sequence that goes to zero, but has -1, everywhere then. — Jorge Ramos, Jun 25 '14 at 03:49
Related: http://math.stackexchange.com/questions/480931/proving-sum-n-geq-1-fraca-n1-a-n-diverges-provided-sum-a-n-diverg . — Clarinetist, Jun 25 '14 at 03:58
@JorgeRamos You can't have $(a_n)=(r_1,r_2,-1,r_4,-1,\ldots)$ because then some $n$ has $a_n=-1$ and so $-1/0$, an element of the partial sum, is undefined. The question is undefined if any $a_n=-1$. — , Jun 25 '14 at 04:02
That was my question, more like a philosophical question, because the definition of converge of series says that if the limit exist and is finite, but that would imply that if the limit doesn't exist the series not convergence. — Jorge Ramos, Jun 25 '14 at 04:07

score 3 · Accepted Answer · answered Jun 25 '14 at 09:24

This is not always true. Let us take $a_n=\frac{1}{n}+\frac{(-1)^n}{\sqrt{n}}$. The series $\sum a_n$ obviously diverges, but $$\frac{a_n}{1+a_n}=a_n-a_n^2+O(a_n^3)=\frac{1}{n}+\frac{(-1)^n}{\sqrt{n}}-\left(\frac{1}{n}+\frac{(-1)^n}{\sqrt{n}}\right)^2+O\left(\frac{1}{n^{3/2}}\right)=\frac{(-1)^n}{\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right),$$ so $\sum \frac{a_n}{1+a_n}$ converges.

score 0 · Answer 2 · answered Jun 25 '14 at 04:18

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If $a_n < 1$, $\frac{a_n}{1+a_n} > \frac{a_n}{2} $. If $a_n \ge 1$, $\frac{a_n}{1+a_n} \ge \frac{1}{2} $. The sum of either of these diverge.

answered Jun 25 '14 at 04:18

marty cohen

107,799

If a_n <0 , a_n/(1+a_n) > a_n/2 is not always true. I think you are assuming a_n >0 – user128766 Jun 25 '14 at 04:27

Does the transformation $a_n \leadsto \frac{a_n}{1+a_n}$ unconditionally preserve divergence of a series?

2 Answers2