TITLE: Convergence/divergence of $\sum\frac{a_n}{1+na_n}$ when $a_n\geq0$ and $\sum a_n$ diverges
QUESTION [23 upvotes]: A question from Rudin (Principles) Chapter 3:

Let $a_n\geq0$ and $\sum a_n$ diverges. What can be said about convergence/divergence of $\sum\frac{a_n}{1+na_n}$?

This one is being recalcitrant. Given that $x>y$ implies $\frac{x}{1+nx}>\frac{y}{1+ny}$ and when $a_n=1/n\log n$ the sum in question diverges, it seems plausible that in general the sum will always diverge, but I can't get a proof out. If it does diverge, it does so pretty slowly as
$$\frac{a_n}{1+na_n}=\frac{1}{n}-\frac{1}{n+n^2a_n}\leq\frac{1}{n}.$$

REPLY [20 votes]: Let $A=\sum\limits_n a_n$ denote a series with nonnegative terms, and $B=\sum\limits_nb_n$ with $b_n=\frac{a_n}{1+na_n}$. Of course, $b_n\leqslant a_n$ hence, if $A$ converges, so does $B$. But, if $A$ diverges, $B$ may diverge or converge.
Example 1: Assume that $a_n=\frac1n$ for every $n\geqslant1$. Then, $b_n=\frac1{2n}$ for every $n\geqslant1$ hence $A$ and $B$ both diverge.
Example 2: Assume that $a_n=1$ for every $n$ in some subset $S$ of the integers, and $a_n=0$ otherwise. Then, $A$ diverges if and only if $S$ is infinite and $B$ diverges if and only if $\sum\limits_{n\in S}\frac1{n+1}$ diverges. If $S$ is the set of squares of integers, $A$ diverges and $B$ converges.
Example 2': If one needs that every $a_n$ is positive, assume that $a_n=1$ for every $n$ in $S$ the set of squares of integers and $a_n=\frac1{n^2}$ otherwise. Then,  $A$ diverges and $B$ converges.