Give an example of divergent series $\sum{a_n}$ with $(a_n)$ decreasing and such that $\lim{na_n} =0$ Same as title, 

Example of divergent series $\sum{a_n}$ with $(a_n)$ decreasing and such that $\lim{na_n} =0$

I can't seem to think a series to match the criteria. The series I am trying, $\lim{na_n}=1$.
 A: How about $$
a_{n}=\frac{1}{n\log n}?
$$
A: Let $H_n:=1+\frac12+\cdots+\frac1{n},\; s_n:=\sqrt H_n,\; a_n:=s_n-s_{n-1}$. But the Harmonic series $H_n$ diverges, so does $s_n$ and  $\;n(H_n-H_{n-1})=1,\;$ so $na_n=n(s_n-s_{n-1})=1/(\sqrt{H_n}+\sqrt{H_{n-1}})$ goes to $0$.
A: The question is "give a counterexample to the converse of 3.23 in Rudin's Principles".
$~a_n = 1/n~$ is a sufficient example that doesn't use logarithm by 3.28.
the thumbnail sketch of why 1/n diverges (3.27) is based on observing that convergence or divergence of the sequence of a monotonically decreasing series a_n can be determined by subsequences such as $~a_1, ~a_2, ~a_4, ~a_8, ~\cdots ~$. $~\sum a_n~$ converges iff $~a_1 + 2a_2 + 4a_4 + \cdots ~$ converges.  see that this is true because partial sums of the former are bounded from above by partial sums of the latter and from below by halves of partial sums of the latter using associativity to compare chunks of elements of the original sequence to the sequence $~a_l = a_{n^l}~$.
since converge is equivalent to bounded partial sums in this scenario, you're done and without the need to invoke the inverse of power series before you know what power series are in the first place!
