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Suppose $a_n>0$, $s_n=a_1+ \dots+ a_n$, and $\sum a_n $diverges.

I need to prove that $\sum \frac{a_n}{1+a_n}$ diverges.

My attempt: We have $\forall n \in \mathbb{N}^*, a_n>0$

$\forall n \in \mathbb{N}^*, \frac{a_n}{1+a_n}-a_n\\= \frac{a_n^2}{1+a_n} \sim {a_n}^2 $

Since $\sum a_n$ diverges, then $\sum {a_n}^2$ diverges (is this true ?) By the equivalence theorem, we can deduce that the series $\sum\big(\frac{a_n}{1+a_n}-a_n\big) $ diverges.

Since $\forall n \in \mathbb{N}^*, \frac{a_n}{1+a_n}=\big(\frac{a_n}{1+a_n}-a_n\big)+a_n$

Then: $\sum\frac{a_n}{1+a_n}$ diverges

Can someone verify my proof ?

amir
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1 Answers1

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HINT: argue for a contradiction, and assume $\sum\frac{a_n}{1 + a_n}$ converges. Then, it must be the case that $\frac{a_n}{1 + a_n} \to 0$.

Use this to show that $a_n$ is eventually bounded by $1$, then use this to show that for some $N$, if $n>N$ then

$$ a_n \leq C\frac{a_n}{1 + a_n} $$

and finish using the comparison test.

BaronVT
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  • Thank you but I have a question though.. if $\frac{1}{1+\frac{1}{a_n}} \to 0$ does that imply that $a_n \to 0$ ? – amir Oct 15 '13 at 18:24
  • Well, note that $\frac{1}{1 + \frac{1}{a_n}}\frac{a_n}{a_n} = \frac{a_n}{1+a_n}$ so the assertion is the same. Though after some thought, the fact that these $\to 0$ doesn't imply $a_n \to 0$ as obviously as I though (if at all). Though it will be the case that $a_n/(1+a_n) \to 0$ will show that $a_n \leq 1$ eventually, which is enough to carry out the rest of the proof. – BaronVT Oct 15 '13 at 18:43