I'd like to show $\sum_n a_n$ converges if and only if $\sum_n \frac{a_n}{1+a_n}$ converges. Where each $a_n$ is a sequence of positive real numbers.
The first side is trivial since $a_n > a_n / (1 + a_n)$ for all $n$, so the result follows by comparison.
I'm having trouble rigorously arguing $\sum_n \frac{a_n}{1+a_n}$ converging implies the convergence of $\sum_n a_n$.
Because $\sum\frac{a_n}{1+a_n}$ converges $a_n\to0$ so $a_n\le1$ for $n$ large enough. In this case $\frac{2 a_n}{1+a_n}\ge a_n$ so you can use comparison again.
Actually, to see what's happening, just consider $A = \sum a_n$ and $B = \sum f(a_n)$. As $A$ converges implies $a_n \rightarrow 0$, we only need to look at the limit behaviour of $f$ around $0$. Obviously $f(0)=0$ is a pre-requisite.
If $f$ is differentiable in $0$, we have $\displaystyle |f(a_n)| \sim |f'(0)| |a_n| = O(|a_n|)$ when $n \to \infty$, so $A$ converges $\implies B$ converges.
As the case of $a_n=n$ and $f(x) = x^2$ shows, the case where $f'(0)=0$ is problematic since $f^{-1}$ wouldn't be differentiable in $0$ and convergence of $B$ does not imply convergence of $A$.
If $f'(0) \not= 0$, we know that $g=f^{-1}$ will be diff in $0$ with $g'(0) = 1/f'(0)$n which in turn implies $B$ converges $\implies A $ converges.
