TITLE: Convergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$?
QUESTION [9 upvotes]: I need to prove the convergence/divergence of the series $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$ based on the convergence/divergence of the series $\sum_{n=1}^{\infty }a_{n}$. It is given that $a_{n}> 0$, $\forall n\in \mathbb{N}$
If the series $\sum_{n=1}^{\infty }a_{n}$ is convergent, then from $\frac{a_{n}}{1+na_{n}}< a_{n}$ and the comparison test, we conclude that $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$ is convergent. However, if the series $\sum_{n=1}^{\infty }a_{n}$ is divergent, I have no idea how to prove the convergence/divergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$. It is definitely divergent (just take $a_{n}=\frac{1}{n}$), but I have no clue how to prove it.

REPLY [5 votes]: Another easy example for the convergence of $ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}} $ even though $ \sum_{n=1}^{\infty}a_{n} $ is divergent:
Define $ a_n $  as follows.
For each $ n\in \mathbb{N} $,
\begin{equation*}
 a_{n} = \begin{cases}
               1               & \text{; if } n=k^2 \text{ for some  }k\in \mathbb{N}\text{ (i.e. n is a square number) }\\\\
               0               & \text{; otherwise }.\\
\end{cases}
\end{equation*} 
Then obviously $$ \sum_{n=1}^{\infty}a_{n} \text{ is divergent.}$$
Notice that $$ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}}= \sum_{k=1}^{\infty}\dfrac{1}{1+k^2}\leq \sum_{k=1}^{\infty}\dfrac{1}{k^2}. $$Therefore $$ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}}\text{ is convergent.}$$