Convergence of non-negative series 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?
 A: 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
$$\sum_{n=1}^\infty x_n\le(1+x_N)\sum_{n=1}^\infty{x_n\over1+x_n}$$
A: 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 
$$x_n \leqslant C\frac{x_n}{1+x_n}$$ 
A: 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.
A: $$\begin{align}
\sum_{n=0}^\infty \frac{x_n}{1+x_n} < \infty \implies & 
\exists N
\left[ \forall n \ge N 
\left( 
\frac{x_n}{1+x_n} < \frac12 \iff 
x_n < 1 \implies x_n < \frac{2x_n}{1+x_n}\right)\right]\\
\implies & \exists N \left[ \sum_{n=N}^\infty x_n \le 2 \sum_{n=N}^\infty\frac{x_n}{1+x_n} 
\le 2\sum_{n=0}^\infty \frac{x_n}{1+x_n} < \infty
\right]\\
\implies & \sum_{n=0}^\infty x_n = \left(\sum_{n=0}^{N-1} + \sum_{n=N}^\infty\right) x_n < \infty
\end{align}
$$
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 $\sum_{n=0}^\infty x_n$.
