Let $x_n\ge 0,\forall n\in\Bbb N$. Prove that $\sum x_n <\infty\iff\sum \frac{x_n}{1+x_n}<\infty$

Check my proof please.

First part $(\implies)$ Because $x_n$ are non-negative terms then the series $\sum x_n$ and $\sum \frac{x_n}{1+x_n}$ increases. If $\sum x_n$ does not diverge to infinity then is bounded above, and then converges because the series is increasing.

Then I want to show that if $\sum x_n$ converges then $\sum\frac{x_n}{1+x_n}$ converges too.

Suppose that $\sum x_n\to s$. It is enough to observe that if $x_k\ge 0$ then $\frac{x_k}{1+x_k}\le x_k$ for any $k$ then

$$\sum\frac{x_n}{1+x_n}\le\sum x_n\to s$$

Because $\sum\frac{x_n}{1+x_n}$ is bounded above by $s$ and is increasing then converges.$\Box$

Second part $(\impliedby)$ I will prove the contrapositive, i.e.

$$\left(\sum\frac{x_n}{1+x_n}<\infty\implies\sum x_n<\infty\right)\iff\left( \sum x_n\to\infty\implies \sum\frac{x_n}{1+x_n}\to\infty\right)$$

Observe that the above holds because $x_n\ge 0$.

  1. If $(x_n)\to \infty$ then $(\frac{x_n}{1+x_n})\to 1\neq 0$ and then $\sum\frac{x_n}{1+x_n}\to \infty$.

  2. If $(x_n)$ does not diverges to infinity then $\sup\{x_n:n\in\Bbb N\}=M$ is finite, and we can observe that

$$\left(\sum\frac{x_n}{1+x_n}\ge\frac{\sum x_n}{1+M}\right)\;\land\;\left(\sum x_n\to\infty\right)\implies \sum\frac{x_n}{1+x_n}\to\infty$$

And with this conclude the second part of the proof.$\Box$

  • 2
    $\begingroup$ I'd say it in much less words. But I find no mistake. $\endgroup$ – ajotatxe Oct 25 '16 at 16:30
  • $\begingroup$ @ajotatxe si quieres añade tu versión, para complementar. En el libro que estoy siguiendo de momento no hay ratio test, root test o test integral. A lo más sólo algunas definiciones y el test de convergencia de series alternas. $\endgroup$ – Masacroso Oct 25 '16 at 16:35

We have

$\forall n\in \mathbb N $


with $$\epsilon(n)=-\frac{x_n}{1+x_n}$$

which goes to $0$ when $n \to\infty$.


the general terms of both series are positive and equivalent near $\infty.$

or, in other terms, for large enough $n$, $0\leq\frac{1}{2}x_n\leq y_n \leq \frac{3}{2}x_n.$

so, they converge together or diverge.


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