If $\sum\limits_{n=1}^{\infty}a_{n}$ diverges, does $\sum\limits_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge? Suppose $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges. Does $\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge?
 A: Eric has proven that the divergence of
$$
\sum_{n=1}^\infty a_n\tag{1}
$$
does not imply the divergence of
$$
\sum\limits_{n=1}^\infty\frac{a_n}{1+na_n}\tag{2}
$$
even if $a_n\ge0$.

However, if $a_n$ decreases monotonically to $0$, then $(2)$ also diverges.
First, note that $\displaystyle\frac{a_n}{1+na_n}$ also decreases monotonically to $0$ since
$$
\frac{a_{n+1}}{1+(n+1)a_{n+1}}=\frac1{1/a_{n+1}+(n+1)}\le\frac1{1/a_n+n}=\frac{a_n}{1+na_n}\tag{3}
$$
Next, note that for $x\ge0$,
$$
\frac{x}{1+x}\ge\frac12\min\left(x,1\right)\tag{4}
$$
Setting $x=na_n$ and dividing by $n$, $(4)$ becomes
$$
\frac{a_n}{1+na_n}\ge\frac12\min\left(a_n,\frac1n\right)\tag{5}
$$
Now, there are two cases: 


*

*$a_n\ge\frac1n$ infinitely often

*there is an $N$ so that for $n\ge N\Rightarrow a_n\lt\frac1n$.
In case 1, we can generate a sequence $n_k$ so that $n_{k+1}\ge2n_k$ and $a_{n_k}\ge\frac1{n_k}$. Then
$$
\begin{align}
\sum_{n=n_k+1}^{n_{k+1}}\frac{a_n}{1+na_n}
&\ge(n_{k+1}-n_k)\frac{a_{n_{k+1}}}{1+n_{k+1}a_{n_{k+1}}}\\
&\ge(n_{k+1}-n_k)\frac1{2n_{k+1}}\\
&\ge\frac14\tag{6}
\end{align}
$$
Since we can find infinitely many such $k_n$, we have that $(2)$ diverges.
In case 2, for $n\ge N$, we have $\frac{a_n}{1+na_n}\ge\frac{a_n}{2}$. Therefore, since $(1)$ diverges, $(2)$ also diverges.
A: The series does not necessarily diverge.  For example, let $a_n$ be the indicator function of the squares, that is the function which is $1$ when $n$ is a the square of an integer, and $0$ otherwise.**   Then the series $\sum_{n=1}^\infty a_n$ diverges, since it fails the divergence test, yet $$\sum_{n=1}^\infty \frac{a_n}{1+na_n}$$ converges by comparison to $\sum_{n=1}^\infty \frac{1}{n^2}$. 
Remark: This is problem $11$ $(d)$ from chapter $3$ of Rudin's "Principles of Mathematical Analysis Third Edition."  I'll add that I think this is a great problem.  The solution is strikingly simple, yet most solvers take a very long time to arrive at it, and erroneously try to prove divergence.
** If you require that $a_n>0$ let $a_n=2^{-n}$ on the non-squares to arrive at the same end result.
