One thing I notice we can do is say $\frac{x_n}{1+x_n} = 1 - \frac{1}{1+x_n}$ but I am not sure how this helps. By Cauchy criteria we can also say there exists $N \in \mathbb{N}$ such that $|\sum_{n=N+1}^{\infty} \frac{x_n}{x_n+1}| < \epsilon.$ Could I use these here? I see the direction from right to left is limit comparison test, but what about left to right?
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As $$0 \le \frac{x_n}{1+x_n}\le x_n,\\ \sum {x_n}<\infty\Rightarrow \sum \frac{x_n}{1+x_n}<\infty $$
Now suppose that $\sum \frac{x_n}{1+x_n}<\infty $.
$\frac{x_n}{1+x_n}\to 0$, and so $x_n\to 0$, so there is a $N$ such as $ n>N\Rightarrow x_n<1 $ and then$$ \frac{x_n}{1+x_n}>\frac {x_n}2 $$so $$ \sum \frac{x_n}{1+x_n}<\infty \Rightarrow \sum {x_n}<\infty $$
mookid
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1This looks good. I know its seems obvious, but how can can we be rigorous about $\frac{x_n}{1+x_n} \to 0$ implies $x_n \to 0$? Proof by contradiction? – user77404 Mar 12 '14 at 01:37
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2if $\frac{x_n}{1+x_n} < r$ then $x_n < r + rx_n$ then $x_n < \frac{r}{1-r}$. – mookid Mar 12 '14 at 01:38
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For the other way:
Hint: If $\sum\frac{x_n}{1+x_n}$ converges, then in particular $x_n$ must converge to $0$, and in particular there exists a constant $C\geq0$ such that $x_n\leq C$. Then $$\sum\frac{x_n}{1+x_n}\geq\sum\frac{x_n}{1+C}.$$
rfauffar
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if $x_n\to 0$.
– Pedro Mar 12 '14 at 01:34