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Let $x_n$ be a sequence of non-negative numbers. I want to show

$\sum_{n=1}^\infty x_n<\infty \Longleftrightarrow \sum_{n=1}^\infty \frac{x_n}{1+x_n}<\infty$

One direction is easy:

$x_n\geq\frac{x_n}{1+x_n}\;\;\forall n$, therefore $\sum_{n=1}^\infty x_n<\infty \Longrightarrow \sum_{n=1}^\infty \frac{x_n}{1+x_n}<\infty$

Any hints for the other direction?

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Any convergent series of positive numbers has a maximum element. Therefore, if $\sum{x_n\over1+x_n}$ converges, then there exists an $N$ such that

$${x_n\over1+x_n}\le{x_N\over1+x_N}\quad\text{for all }n$$

From this we see that

$${x_n\over1+x_n}=x_n\left(1-{x_n\over1+x_n}\right)\ge x_n\left(1-{x_N\over1+x_N}\right)=x_n\left(1\over1+x_N\right)$$

and therefore

$$\sum_{n=1}^\infty x_n\le(1+x_N)\sum_{n=1}^\infty{x_n\over1+x_n}$$

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Hint:

If $\sum \frac{x_n }{ 1 + x_n}$ converges, then $1 - \frac{1}{1+x_n}= \frac{x_n}{1+x_n} \to 0$ which implies $x_n \to 0$.

Now show for sufficiently large $n$ there exists $C$ such that

$$x_n \leqslant C\frac{x_n}{1+x_n}$$

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We use the limit comparison test for positive series.

$\sum x_n$ converges

$\implies \lim_{n\to+\infty} x_n=0$

$\implies x_n\sim \frac{x_n}{1+x_n} \;(n\to+\infty)$

$\implies \sum \frac{x_n}{1+x_n}$ converges.

AND

$\sum \frac{x_n}{1+x_n} $converges

$\implies \lim_{n\to+\infty} x_n=0$

$\implies \frac{x_n}{1+x_n}\sim x_n \; (n\to +\infty)$

$\implies \sum x_n$ converges.

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$$\begin{align} \sum_{n=0}^\infty \frac{x_n}{1+x_n} < \infty \implies & \exists N \left[ \forall n \ge N \left( \frac{x_n}{1+x_n} < \frac12 \iff x_n < 1 \implies x_n < \frac{2x_n}{1+x_n}\right)\right]\\ \implies & \exists N \left[ \sum_{n=N}^\infty x_n \le 2 \sum_{n=N}^\infty\frac{x_n}{1+x_n} \le 2\sum_{n=0}^\infty \frac{x_n}{1+x_n} < \infty \right]\\ \implies & \sum_{n=0}^\infty x_n = \left(\sum_{n=0}^{N-1} + \sum_{n=N}^\infty\right) x_n < \infty \end{align} $$ In human words, $\sum_{n=0}^\infty \frac{x_n}{1+x_n}$ is finite implies after some $N$, all $\frac{x_n}{1+x_n} < \frac12$ and this leads to a bound $x_n \le \frac{2x_n}{1+x_n}$ for $n \ge N$. You can then use this to control the tail of the sum $\sum_{n=0}^\infty x_n$.

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