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Ive tried to prove that the partial sums of $\dfrac{a_n}{a_{n}+1}$ are bounded, but besides of writing it as the sum $1-\dfrac{1}{a_{n}+1}$, I cannot imagine how to do it...

Diego
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    Well as each term $0< a_n/(a_n+1) < a_n/1 = a_n$ then the sum of the terms must be $0 < \sum a_n/(a_n+1) < \sum a_n$ so ... – fleablood Feb 19 '16 at 04:02
  • Oh, it was easier than I thought. So I guess we can generalize replacing $a_n +1$ by $b_n$, with the condition that $b_n>a_n$ from some $n$ onwards – Diego Feb 19 '16 at 04:16
  • You should have a theorem in your aresenal that if $a_n \rightarrow a$ and $b_n \rightarrow b$ and $b_n < c_n < a_n$ and monotonic (so it doesn't "bounce" between a and b) then $c_n $ converges. – fleablood Feb 19 '16 at 04:25

2 Answers2

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Note this

$$\lim_{n\to \infty} \dfrac{a_n}{a_n(a_{n}+1) }=1$$

Which implies that the series converges since $\sum_n a_n$ does (by comparison theory test). See here.

Community
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Mhenni Benghorbal
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  • I think yours is best. It doesn't even use that $a_n>0$ – Diego Feb 19 '16 at 16:21
  • @Diego:In fact the comparison theorem requires both series to be $>0$ but it is easy and efficient! – Mhenni Benghorbal Feb 19 '16 at 20:49
  • I mean both series have positive terms. – Mhenni Benghorbal Feb 19 '16 at 21:31
  • Yeah,I meant that it could be $a_n<0$ too. – Diego Feb 19 '16 at 23:04
  • @Diego I am not sure you got the point the first time so let me reiterate: the result is false if one does not assume that $a_n>0$. – Did Feb 21 '16 at 11:34
  • @Did As far as I see, the limit comparision theorem holds too if the sequences to compare are both negative from some $N$ onwards. Furthermore, it holds even if one is negative and the other positive from some $N$. In particular, if $-1<a_n<0$ from some $M$, then $\dfrac{a_n}{a_{n}+1}$ is negative after that $M$, too. So the result follows. – Diego Feb 21 '16 at 16:59
  • @Diego This is not the point: if $(a_n)$ is real valued, the signs can alternate--and then the result fails. – Did Feb 21 '16 at 17:06
  • Indeed, it's not necessary to add $-1<a_n<0$ as an additional hypothesis, since it follows from the facts that $a_n$ converges to 0 and $a_n<0$ – Diego Feb 21 '16 at 17:08
  • Yeah, I agree that if $a_n$ alternates the result does not follow. – Diego Feb 21 '16 at 17:09
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We have $a_n>0$, then $a_n+1>1$ and $$ 0<\dfrac{a_n}{a_{n}+1}<a_n $$ giving $$ 0<\sum\dfrac{a_n}{a_{n}+1}<\sum a_n. $$ The series is convergent.

Olivier Oloa
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