Prob. 11 (d), Chap. 3 in Baby Rudin: Given $a_n > 0$, is this condition also sufficient for divergence of $\sum \frac{a_n}{1+na_n}$?

• 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.