Can someone please verify my proof sketch?
Suppose $a_n \geq 0$, and $\sum a_n$ diverges, and $\lim a_n = 0$. Show that $\displaystyle{\sum \frac{a_n}{1+a_n}}$ diverges.
Let $\epsilon > 0$. Then,
\begin{eqnarray}
\exists N \in \mathbb{N} \ \ s.t. \ \ \forall n > N, 0 \leq a_n < \epsilon
\end{eqnarray}
This implies that
\begin{eqnarray}
\forall n > N, \displaystyle{\frac{a_n}{1+\epsilon} < \frac{a_n}{1+a_n}}
\end{eqnarray}
But then,
\begin{eqnarray}
\displaystyle{\sum\limits_{k=N+1}^\infty \frac{a_k}{1+\epsilon} < \sum\limits_{k=N+1}^\infty \frac{a_n}{1+a_n}}
\end{eqnarray}
Since $\sum a_n$ diverges, so does $\displaystyle{\sum\limits_{k=N+1}^\infty \frac{a_k}{1+\epsilon}}$.
By the comparison test, $\displaystyle{\sum\limits_{k=N+1}^\infty \frac{a_n}{1+a_n}}$ also diverges.
Best Answer
If you use Limit comparison test then you get $$\lim\limits_{n\to\infty}\frac{\frac{a_n}{1+a_n}}{a_n}=\lim\limits_{n\to\infty}\frac{1}{1+a_n}=1.$$ This means that the series $\sum a_n$ and $\sum\frac{a_n}{1+a_n}$ have same character. So $\sum\frac{a_n}{1+a_n}$ diverges.