If $\sum a_n$ diverges then $\sum \frac{a_n}{\ln n}$ diverges If $\sum a_n$ diverges does $\sum \frac{a_n}{\ln n}$ necessarily diverge for $a_n>0$?
I tried $a_n=\frac{\ln n}{n^2}$ to try bag an easy counterexample but it turns out $\sum a_n$ converges(even things like $\frac{\ln n}{n^{1.0001}} $ converge so Id suspect this entire class converges). I also tried things like $a_n=\ln \ln \ln \ln n$ but even for these, both sums diverge. Don't tell me its true?
 A: No this is not true.
For example $\sum\frac{1}{n\cdot \ln n}$ diverges, but $\sum\frac{\frac{1}{n\cdot \ln n}}{\ln n}=\sum\frac{1}{n\cdot (\ln n)^2}$ converges by the integral test.
A: A counter example would be $\displaystyle\sum_{n=2}^\infty\frac1{n\log(n)}$. It diverges, but the series $\displaystyle\sum_{n=2}^\infty\frac1{n\log^2(n)}$ converges.
A: If you are willing to chuck any and all monotonicity out the window, then you can prove something much stronger:
Claim: Given any sequence $(b_n)$ of positive reals converging to zero, one can find a sequence $(a_n)$ of positive reals such that:

*

*$(a_n)$ does not converge to zero, so that $\sum a_n$ diverges in particular.

*$\sum a_n b_n$ converges.

Proof: Choose an increasing sequence of positive integers $n_1<n_2<n_3<\ldots$ such that $b_{n_k} < \frac{1}{2^k}$ and define
\begin{align*}a_n = \begin{cases}
1 & \text{ if $n=n_k$ for some $k$ } \\
\frac{1}{2^n b_n} & \text{ otherwise. }
\end{cases} && && && && && \square\end{align*}
Remark: Of course if $a_n$ nonnegative is allowed, then you can just use $0$ in place of $\frac{1}{2^nb_n}$ above for an even easier answer.
A: Other answers are enough, just an observation on how to derive a counterexample: using Abel summation formula, aka discrete integration by parts, the general term is
$$\sim \sum A_n (1/ \log(x)) '_{x = n} =-  \sum A_n (1/n \log(n) ^2 ) $$
Here $A_n = \sum_{k \le n} a_k$. If $A_n \sim \log \log(n) \to \infty, $ you are still left with a
$$\sum \frac{\log \log(n) }{n \log(n) ^2}$$
that converges. If  $A_n \sim \log\log(n) $ we expect
$$a_n \sim (\log \log(n)) ' = \frac{1}{n\log(n) }$$
which is the counterexample that has been provided.
