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Although this question has been posted here many times, I am posting it again in order to get my proof verified.

Firstly, we break up the index set of $(u_n)_n$ into a partition of two sets $A$ and $B$ such that:

$A= \{n\in \mathbb{N}:u_n<1\}$ and $B=\{n\in \mathbb{N}: u_n \ge 1 \}$.

Now, $u_n<1 \implies u_n +1<2 \implies \displaystyle\frac{u_n}{2}<\frac{u_n}{1+u_n} $. Taking summation over $A \ \ $,

$\frac{1}{2}\displaystyle\sum_{n \in A}u_n < \sum_{n \in A} \frac{u_n}{1+u_n} ...(1)$

Again, $u_n \geq 1 \implies 2u_n\geq1+ u_n \implies \displaystyle\frac{u_n}{1+u_n} \ge \frac{1}{2} $

The sum over $B$ becomes $\frac{1}{2}\displaystyle |B| \le \sum_{n \in B} \frac{u_n}{1+u_n} ...(2)$ [$|B|$ denotes the number of elements in $B$].

Since $A$ and $B$ form a partition of $\mathbb{N}$, then either of them (or both) must be an infinite set.

Three cases may arise:

\begin{cases} A \mathbb{\ is \ infinite \ but} B \mathbb{ \ is \ finite} \\ B \mathbb{\ is \ infinite \ but} A \mathbb{ \ is \ finite} \\ \mathbb {both \ are \ infinite} \end{cases}

In the first, second and third cases, the left hand sides of $1$, $2$, & $1 { \ \mathbb{and} \ } 2$ diverges respectively. In any case, $\sum \frac{u_n}{1+u_n}$ diverges.

Please verify this method. It will be of immense help for me.

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  • $\begingroup$ I found out later (after posting it) that this proof almost goes along the same line: math.stackexchange.com/a/2794447/389992 $\endgroup$ – Subhasis Biswas May 4 at 6:51
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    $\begingroup$ if $\sum_n \frac{u_n}{1+u_n} < \infty$, then $\frac{u_n}{1+u_n} \to 0$, so $u_n \to 0$, so $\frac{u_n}{1+u_n} \ge \frac{1}{2}u_n$ for large $n$, so $\sum_n \frac{u_n}{1+u_n}$ diverges by comparison test $\endgroup$ – mathworker21 May 4 at 6:52
  • $\begingroup$ That's a bit simpler... $\endgroup$ – Subhasis Biswas May 4 at 6:58
  • $\begingroup$ More solutins e.g. here: math.stackexchange.com/q/131678/42969. $\endgroup$ – Martin R May 4 at 9:26
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I'm transferring @mathworker21's solution into an answer, so I'll make this CW.

If $\sum \dfrac{u_n}{1+u_n} < \infty$, then $\dfrac{u_n}{1+u_n} \to 0$, so $u_n \to 0$, so $\dfrac{u_n}{1+u_n} \ge \frac12 u_n$ for $n$ sufficienty large, so $\sum \dfrac{u_n}{1+u_n}$ diverges by comparison test.

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  • $\begingroup$ How do you make CW @GNUSupporter 8964民主女神 地下教會 $\endgroup$ – Aqua May 4 at 15:38
  • $\begingroup$ @MariaMazur There's a checkbox at the right bottom corner. Just tick that will do. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 4 at 15:40
  • $\begingroup$ aha, thanks @GNUSupporter8964民主女神地下教會 $\endgroup$ – Aqua May 4 at 15:41
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(Proof 1). (i). If $\lim_{n\to \infty}u_n=0$ then $u_n<1/2$ for all but finitely many $n.$ So $u_n/(1 +u_n)\ge u_n/(1+1/2)=(2/3)u_n$ for all but finitely many $n.$

(ii). If $\neg (\lim_{n\to \infty}u_n=0\}$ then there exists $r>0$ such that $u_n>r$ for infinitely many $n.$ And $u_n>r>0 \implies u_n/(1+u_n)>r/(1+r).$ So $\neg (\lim_{n\to \infty}u_n/(1+u_n)=0).$

(Proof 2). Let $v_n=u_n/(1+u_n),$ so $u_n=v_n/(1-v_n).$ The Q is equivalent to: If $0\le v_n<1$ for all $n$ and $\sum_nv_n$ converges then $\sum_nv_n/(1-v_n)$ converges. Proof: $\lim_{n\to \infty}v_n=0$ so for all but finitely many $n$ we have $v_n<1/2.$ So $v_n/(1-v_n)\le v_n/(1-1/2)=2v_n $ for all but finitely many $n$.

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