If $\{a_n\}$ is not summable, neither is $\left\{ {\frac{{{a_n}}}{{1 + {a_n}}}} \right\}$

Suppose that $a_n\geq 0 (n\in \mathbb{N})$ and $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+a_n} \; \text{is convergent}$. Then $$ \lim_{n\rightarrow\infty}\frac{a_n}{1+a_n}=0. $$ It implies that $\displaystyle\lim_{n\rightarrow\infty}a_n=0$ and $$ \lim_{n\rightarrow\infty}\frac{a_n}{\frac{a_n}{1+a_n}}=\lim_{n\rightarrow\infty}(1+a_n)=1. $$ Since $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+a_n} \; \text{is convergent}$, we have $\displaystyle\sum_{n=1}^{\infty}a_n \; \text{is convergent}$

Assume that $$ \sum\limits_{n=1}^\infty \frac{a_n}{1+a_n}<+\infty $$ then $$ \lim\limits_{n\to\infty}\frac{a_n}{1+a_n}=0 $$ This implies $\displaystyle\lim\limits_{n\to\infty}a_n=0$ and since $a_n\geq 0$ for all $n$ we get $N\in\mathbb{N}$ such that $|a_n|<1$ for all $n>N$. For such $n$ we have $$ a_n<\frac{2a_n}{1+a_n} $$ As the consequence $$ \sum\limits_{n=N}^\infty a_n< 2\sum\limits_{n=N}^\infty \frac{a_n}{1+a_n}<+\infty $$ The rest is clear.