Given $\Sigma a_n$ diverges show that $\Sigma \frac{a_n}{1+a_n}$ diverges.

Question

Intuitively speaking, I first thought that if the series $\Sigma a_n$ is divergent then

$$\lim_{n \to \infty} a_n \ne 0$$

therefore it was clear that $\Sigma \frac{a_n}{1+a_n} $ would be divergent, but when I thought about it there are cases where the limit of the sequence does approach to $0$ and yet diverge, like the harmonic series.

Then I tried to go with since the sequence diverges, the series is not Cauchy (I m not even 100% sure if this is true but I tried)

$$|\sum_{i = m}^{n} a_n| \gt \epsilon$$

and derive the other series to not be Cauchy as well, only to not being able to reach.

I appreciate all the help.

Answer 1

1. If $(a_n)$ is a non-negative sequence, as commented by Panda and David Mitra, the statement is true. This is because, if $\sum_n\frac{a_n}{1+a_n}$ converges, then $\lim\limits_{n\to\infty} a_n=0$, so by comparison test, $\sum_n a_n$ converges, a contradiction.
2. In general, the statement is false. For example, as a similar construction here, let $b_n=\frac{(-1)^n}{\sqrt{n}}$ and let $a_n=\frac{b_n}{1-b_n}$. Then $b_n=\frac{a_n}{1+a_n}$ and $$a_n=b_n+b_n^2+\frac{b_n^3}{1-b_n}.\tag{1}$$ It is easy to see that $\sum_n b_n$ is convergent, $\sum_n b_n^2$ is divergent and $\sum_n \frac{b_n^3}{1-b_n}$ is convergent( because $b_n\to 0$ and $\sum_n |b_n|^3$ is convergent), so from $(1)$ we know that $\sum_n a_n$ is divergent.