Is $\lim_{n\to\infty}\frac{n(a_n-a_{n+1})}{a_{n+1}}=0$ if $\sum_{n\geq 1} a_n/n$ diverges? Consider a sequence $(a_ n)$ of positive and monotonically decreasing terms for which $\sum_{n\geq 1} a_n/n$ diverges and $a_ n\to 0$. I claim that that $$\lim_{n\to\infty}\frac{n(a_n-a_{n+1})}{a_{n+1}}=0.$$ 
I suspect that it's true (at least I can't find a counterexample) but I can't find a clean proof. Is there some theorem or test that might help out?
 A: Try $a_n=\frac{1}{\log (n+1) \prod_{k^2\le n}\left(1+\frac{1}{k^2}\right)}$. Note that:


*

*$a_n$ is monotonically decreasing to $0$

*$a_n$ is bounded below by $\frac{1}{\log (n+1) \prod_{k\ge1}\left(1+\frac{1}{k^2}\right)}=\frac{\pi}{\sinh\pi}\frac{1}{\log(n+1)}$

*$\sum_n\frac{a_n}{n}=\infty$ by comparison with, say, $\int_2^t\frac{1}{x\log x}dx$


Now, note that for $n=k^2$, $\frac{a_n}{a_{n-1}}=\frac{\log(n)}{\log(n+1)(1+\frac{1}{n})}$, whence the listed ratio at $(n-1)$ is:
$$(n-1)(\frac{a_{n-1}}{a_n}-1)=(n-1)\left(\frac{\log(n+1)(1+\frac{1}{n})}{\log(n)}-1\right)$$
Now, notice $\frac{\log(n+1)}{\log(n)}=1+\frac{1}{n\log(n)}+O(\frac{1}{n^2})$, giving the ratio as:
$$\left(n-1\right)\left(\left(1+\frac{1}{n\log(n)}+O(\frac{1}{n^2})\right)\left(1+\frac{1}{n}\right)-1\right)=(n-1)(\frac{1}{n}+O(\frac{1}{n\log(n)}))\to1$$
which offers a counterexample to the claim.
One can also note that for other $n$, the ratio is $n\left(\frac{\log(n+2)}{\log(n+1)}-1\right)=O\left(\frac{1}{\log (n)}\right)\to0$.
