Prob. 11 (d), Chap. 3 in Baby Rudin: Given $a_n > 0$, is this condition also sufficient for divergence of $\sum \frac{a_n}{1+na_n}$? Here's Prob. 11 (d), Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

Suppose $a_n > 0$ and that the series $\sum a_n$ is divergent. Then what can be said about the convergence of $$\sum \frac{a_n}{1+na_n}?$$

I know that if $\left\{ n a_n \right\}$ is bounded above or has a positive lower bound, then this series diverges. Does the converse hold as well? 
Suppose that $\left\{ n a_n \right\}$ is neither bounded above nor has a positive lower bound. Then there is a subsequence $\left\{ n_k a_{n_k} \right\}$ such that $$n_k a_{n_k} \geq k$$ for all $k$. 
And, there is a subsequence $\left\{ m_r a_{m_r} \right\}$ such that $$m_r a_{m_r} < \frac{1}{r}$$ for all $r$. 
What next? 
 A: *

*Answering the Baby Rudin question:
You cannot say anything.
It may be divergent: for instance, take $(a_n)_{n\geq 1}$ to be identically $1$, or even $(a_n)_{n\geq 1}$ to be defined by $a_n = \frac{1}{n}$ (two natural examples).
It may be convergent: for instance, $a_n = \begin{cases} 1 &\text{ if } n=2^k \text{ for some }k\geq 0\\ 0 &\text{ otherwise}\end{cases}$ (you can replace $0$ by $2^{-n}$ if you want to enforce that the sequence be positive).
Since $(a_n)_n$ does not converge to $0$, clearly the series $\sum_n a_n$ diverges. Yet, $\sum_{n=1}^\infty \frac{a_n}{1+n a_n} = \sum_{k=0}^\infty \frac{1}{1+2^k} < \infty$.


*

*Answering the  OP's followup question:
Take $(a_n)_{n\geq 1}$ defined by
$$
a_n = \begin{cases}
2^n & \text{ for even } n\\
\frac{1}{2^n} & \text{ otherwise.}
\end{cases}
$$
Then $\sum_{n=1}^\infty a_n$ clearly diverges, and so does 
$\sum_{n=1}^\infty \frac{a_n}{1+n a_n} \geq \sum_{n=1}^\infty \frac{a_{2n}}{1+2n a_{2n}} = \sum_{n=1}^\infty \frac{1}{\frac{1}{2^{2n}}+2n}$. But $(a_n)_{n\geq 1}$ has neither a finite upper bound nor a positive lower bound.
A: This one is super hard but this is what I have; the claims can be proven but I didn't do that here. Consider this to be messy work instead:
$\textbf{Claim 1}$: If inf $\left\{a_{n}\right\}=0$, then $\left\{a_{n}\right\}$ has a subsequence that converges to $0$.
$\textbf{Claim 2}$: If $a_{n}\longrightarrow 0$, then $\left\{a_{n}\right\}$ has a decreasing subsequence. 
$\textbf{Claim 3}$: $s_{n}=a_{1}+\cdots+a_{n}$ forms a divergent sequence of partial sums, where $a_{n}>0$, then all of it's subsequences are unbounded.
Now, if $\sum a_{n}$ is divergent with inf $\left\{a_{n}\right\}=0, a_{n}>0$, then it has a subsequence that converges to $0$ which in turn has a decreasing subsequence. There is $N$ such that if $n\geq N, s_{n}>1$. If $\left\{s_{n_{k}}\right\}$ 
is the sequence of partial sums whose last term is the smallest (there is such a sequence because we have a decreasing sequence of terms), then for $n_{k}\geq N, \displaystyle  \frac{a_{n_{k}}}{1+n_{k}a_{n_{k}}}\geq  
\frac{a_{n_{k}}}{1+s_{n_{k}}}>\frac{a_{n_{k}}}{2s_{n_{k}}}$. By part(b) and claim 3, the right-most expression has a divergent series and therefore
$\displaystyle \left\{\frac{a_{n_{k}}}{1+n_{k}a_{n_{k}}}\right\}$ has a divergent series. But this series is a subsequence 
of the partial sums 
of $\displaystyle \sum \frac{a_{n}}{1+na_{n}}$, and thus, the original series diverges. 
If $inf \left\{a_{n}\right\}=c>0$ and $a_{n}<1\forall n\Rightarrow \displaystyle \frac{a_{n}}{1+na_{n}}\geq \frac{c}{1+n}$ which forms a divergent series. Finally, if there is a subsequence with $a_{n_{k}}>1\Rightarrow \displaystyle \frac{a_{n_{k}}}{1+na_{n_{k}}}>\frac{a_{n_{k}}}{a_{n_{k}}+na_{n_{k}}}=\frac{1}{1+n}$. 
