Convergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$? 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. 
 A: As you noted, if $a_n=\frac{1}{n}$, then $\sum_{n=1}^\infty \frac{a_n}{1+na_n}=\sum_{n=1}^\infty \frac{1}{2n}$ diverges.
On the other hand, say $a_{2^k}=\frac{1}{2}$ and $a_{n}=2^{-n}$ for $n$ not a power of two. Then
$$\sum_{n=1}^\infty \frac{a_n}{1+na_n} < \sum_{n=1}^\infty \frac{2^{-n}}{1+n2^{-n}} + \sum_{k=1}^\infty \frac{1}{2+2^k} \, ;$$
since both right-hand sums are convergent and every term of the left-hand sum is positive, the left-hand sum also converges.
In both cases, $\sum a_n$ diverges. So merely knowing that $\sum a_n$ diverges is insufficient to tell you about the convergence behavior of $\sum \frac{a_n}{1+na_n}$.
On the other hand, if $a_n$ is decreasing and $\sum a_n$ is divergent, then $\frac{a_n}{1+na_n}>\frac{a_n}{1+s_n}>\frac{a_n}{2s_n}$ for $n$ sufficiently large, where $s_n=\sum_{k=1}^n a_k$. Moreover, I claim that if $\sum a_n$ diverges, then so does $\sum \frac{a_n}{s_n}$. Hence, if you add the condition that $a_n$ is decreasing to the original problem, it will in fact be true that $\sum a_n$ converges if and only if $\sum \frac{a_n}{1+na_n}$ does.
To establish the claim, fix some positive integer $N$ and look at the tail of $\sum \frac{a_n}{s_n}$ beginning at the $N^{\rm th}$ term. For any positive integer $k$, we have
$$\sum_{i=1}^k \frac{a_{N+i}}{s_{N+i}} > \sum_{i=1}^k \frac{a_{N+i}}{s_{N+k}}=\frac{s_{N+k}-s_N}{s_{N+k}}=1-\frac{s_N}{s_{N+k}} \, .$$
Since the sequence $s_n$ is divergent, we can choose $k$ such that $s_{N+k}>2s_N$, and thus $1-\frac{s_N}{s_{N+k}}>\frac{1}{2}$. But $N$ was arbitrary; hence the tails of $\sum \frac{a_n}{s_n}$ don't decrease to zero, and so the sum does not converge.
A: 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 $$ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}}\text{ is convergent.}$$
