I'm writing solutions to exercises in Baby Rudin I think might want to assign students, and I'm having particular difficulty with 11(d) in Chapter 3. Namely, for a sequence $(a_{n})_{n=1}^{\infty}$ of positive reals such that $\sum_{n=1}^{\infty}a_{n}$ diverges, does $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ converge or diverge?
I know that if we take $a_{n} := \frac{1}{n}$, then $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}} =\sum_{n=1}^{\infty}\frac{1}{2n}$ diverges. But I can't seem to come up with an example such that $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ converges while $\sum_{n=1}^{\infty}a_{n}$ diverges. I did a bit of searching online, but the example I came across was $a_{n} = \frac{1}{n(\log n)^{p}}$, where $p > 1$. However, this isn't satisfactory since $\sum_{n=2}^{\infty}\frac{1}{n(\log n)^{p}}$ converges by Theorem 3.29 in Baby Rudin.
Can anyone help me out?