# If $\sum u_n$ be a divergent series of positive reals, then $\sum \frac{u_n}{1+u_n}$ diverges also

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.

• I found out later (after posting it) that this proof almost goes along the same line: math.stackexchange.com/a/2794447/389992 – Subhasis Biswas May 4 at 6:51
• 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 – mathworker21 May 4 at 6:52
• That's a bit simpler... – Subhasis Biswas May 4 at 6:58
• More solutins e.g. here: math.stackexchange.com/q/131678/42969. – Martin R May 4 at 9:26

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.

• How do you make CW @GNUSupporter 8964民主女神 地下教會 – Aqua May 4 at 15:38
• @MariaMazur There's a checkbox at the right bottom corner. Just tick that will do. – GNUSupporter 8964民主女神 地下教會 May 4 at 15:40
• aha, thanks @GNUSupporter8964民主女神地下教會 – Aqua May 4 at 15:41

(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 $$n$$.