Let $ \{a_n\}_{n=1}^\infty $ be a sequence such that $\displaystyle \sum a_n $ that is divergent to $+\infty$.
What can be said about the convergence of $\displaystyle \sum \frac{a_n}{1 + a_n} $?
Any hints, thoughts or leads would be greatly appreciated.
Thanks!
It seems that one assumes that $a_n\geqslant0$ for every $n$. Let $b_n=\frac{a_n}{1+a_n}$. If $a_n\geqslant1$ infinitely often, $b_n\geqslant\frac12$ infinitely often hence $\sum\limits_nb_n$ diverges. Otherwise, $a_n\leqslant1$ for every $n$ large enough, say for every $n\geqslant N$, hence $b_n\geqslant\frac12a_n$ for every $n\geqslant N$ and $\sum\limits_{n\geqslant N}b_n\geqslant\frac12\sum\limits_{n\geqslant N}a_n$, which implies that $\sum\limits_nb_n$ diverges.
Finally, if $a_n\geqslant0$ for every $n$ and if $\sum\limits_na_n$ diverges, then $\sum\limits_n\frac{a_n}{1+a_n}$ diverges.
A classical counterexample when the nonnegativity condition fails is $a_n=\frac{(-1)^{n+1}}{\sqrt{n+1}+(-1)^n}$ for every $n\geqslant1$. Then $\sum\limits_na_n$ diverges because $a_n=\frac{(-1)^{n+1}}{\sqrt{n+1}}+\frac1{n+1}+O\left(\frac1{n^{3/2}}\right)$ but $b_n=\frac{(-1)^{n+1}}{\sqrt{n+1}}$ hence $\sum\limits_nb_n$ converges.
If $0 < a_n/(1+a_n) < 1/2$ then $a_n < 1$, and $0 < a_n < 2 (a_n/(1+a_n))$.
So if $\sum a_n/(1+a_n)$ converges, so does $\sum a_n$.
Because $ \displaystyle \lim_{n \to \infty} a_n = \infty $, it is true that $ \displaystyle\lim_{n \to \infty} \frac{a_n}{1 + a_n} = 1 $. Hence, the terms of the series do not tend to zero so it certainly does not converge.
In response to the edit, here's a brief sketch: Assume, for the sake of contradiction, that $ \sum \frac{a_n}{1 + a_n} $ converges. Hence, its general term $ 1 - \frac{1}{1 + a_n} $ goes to $ 0 $. This means that $ a_n $ goes to $ 0 $. Hence, $ \frac{1}{1 + a_n} = 1 - a_n + O\left(a_n^2\right) $. Hence, $ \sum a_n + O\left(a_n^2\right) $ converges so $ \sum a_n $ converges, a contradiction.
