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I want to prove the following:

Suppose $\sum_\limits{n\in\mathbb{N}}{b_n}$ is divergent. Then $\sum_\limits{n\in\mathbb{N}}{\dfrac{b_n}{1+b_n}}$ is divergent.

I think you have to prove by contradiction and by using Cauchy's criterion. But how to start with that? Or is it possible to use another well-known criterion?

Nohemi Tejeda
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  • don't we have $b_n$ conv/div $\implies 1/b_n$ div/conv – JMP Nov 27 '15 at 12:47
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    no, consider the harmonic series – A Simmons Nov 27 '15 at 12:57
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    @JonMarkPerry: You only have that if $\sum b_n$ converges, then $b_n \to 0$, so $\dfrac 1 {b_n} \not\to 0$, so necessarily $\sum \dfrac 1 {b_n}$ diverges. – Alex M. Nov 27 '15 at 13:44
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    Do we know that $b_n \ge 0 \ \forall n$? – Alex M. Nov 27 '15 at 14:04
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    Put $u_n=\frac{(-1)^n}{\sqrt{n}}$ for $n\geq 1$, Then $u_n$ is a convergent series. If $b_n=\frac{u_n}{1-u_n}$, using $\frac{1}{1-x}=1+x+O(x^2)$, we have $b_n=u_n+\frac{1}{n}+O(1/n^{3/2})$, and we see that the series $b_n$ is divergent. But the series $b_n/(1+b_n)=u_n$ is a convergent series. – Kelenner Nov 27 '15 at 14:23
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    $$b_n=\frac{(-1)^n}{2\sqrt{n}-(-1)^n}\implies\sum_nb_n\ \text{diverges},\ \sum_n\frac{b_n}{1+b_n}\ \text{converges}$$ – Did Nov 27 '15 at 14:24
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    @Did: Your counterexample and the one of Kelenner show that the OP has most probably missed the hypothesis that $b_n \ge 0 \ \forall n$, in which case the question has an answer here, so I voted to close it as a duplicate. – Alex M. Nov 27 '15 at 14:29

2 Answers2

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As Daniel fischer noted this argument only works for the special case of non negative terms $b_i $.

Hint: if the set $\{n: b_n \geq 1\} $ is infinite then you should know what to do. However if that is finite,then try to make use of the following fact:

for every $n $ such that $b_n<1$ we have

$$\frac {b_n}{1+b_n} > \frac {b_n}{2} $$

Amr
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I have an answer when the limit of $b_{n}$ does exist and $b_{n}>0$.

We know that $\sum_\limits{n\in\mathbb{N}}{b_n}$ is divergent so If $\lim_\limits{n\to \infty}b_{n}\neq 0$ so $ \lim_\limits{n\to \infty}\dfrac{b_n}{1+b_n}\neq 0. $ And if $\lim_\limits{n\to \infty}b_{n}= 0$ so we apply the limit test for series $\lim_\limits{n\to \infty}\dfrac{\dfrac{b_n}{1+b_n}}{b_{n}}=1 $ which tells us the $ \sum_\limits{n\in\mathbb{N}}{\dfrac{b_n}{1+b_n}}$ is divergent.

Hamit
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    @Amr I assumed the both condition one time zero and one time non zero. I think it is true. – Hamit Nov 27 '15 at 13:04
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    This answer is completely wrong. First, the limit test only works for series with positive terms, which is not necessarily the context here. Second, if $b_n \not\to 0$, how do you deduce that $\dfrac {b_n} {1 + b_n} \not\to 0$? Third, who says that $b_n$ is supposed to have a limit? – Alex M. Nov 27 '15 at 13:40
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    I said that $b_n$ might not have a limit at all: think about $(-1)^n$ or $\sin n$. – Alex M. Nov 27 '15 at 13:58
  • @AlexM. OK Now I got u. So I edit the answer. – Hamit Nov 27 '15 at 14:04
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    No, it is still wrong: $(-1)^n n$ has no limit, but $\dfrac {(-1)^n n} {1 + (-1)^n n} \to 1$. What you want is to show that if $b_n$ has no limit, then $\dfrac {b_n} {1 + b_n} \not\to 0$; how do you do that? – Alex M. Nov 27 '15 at 14:17
  • @AlexM. I edited the answer again. We should prove that if the $\lim\limits_{n\to \infty}b_{n}$ does not exist then $\lim_\limits{n\to \infty}\dfrac{b_n}{1+b_n}$ does not convergent to zero. – Hamit Nov 27 '15 at 14:30