A necessary condition for series convergence with positive monotonically decreasing terms Suppose that the series $\sum a_n$ is convergent and the terms are positive and decreasing. Is it necessary that $\lim n \log n \, a_n$ exists and 
$$\lim_{n \to \infty} \,\,n \log {n} \,a_n = 0.$$
If either the limit exists and 
$$\lim_{n \to \infty} n \log n \,a_n \neq 0$$ 
or the limit does not exist and 
$$\liminf_{n \to \infty} \,\,n \log n \,a_n > 0,$$
then the series must diverge since $a_n \geqslant C(n \log n)^{-1}$ for all $n$ sufficiently large.
If we relax the monotone condition for $a_n$, we have the example
$$a_n = \begin{cases} \frac{1}{n \log n} &\mbox{if } n = m^2 \\ \frac{1}{n^2} &\mbox{if }n \neq m^2\end{cases}$$
where the series converges but the limit does not exist and 
$$0 = \liminf_{n \to \infty} \,\,n \log n \,a_n < \limsup_{n \to \infty} \,\,n \log n \, a_n = 1.$$
 A: The condition $n\log n \,a_n\to 0$ does not seem to be necessary.
Let us try to find a counterexample of the form
$$a_n=\frac{c_n}{n\log n}\quad,\quad n\geq 2.$$
The sequence $(a_n)$ will be decreasing if and only if $\frac{c_{n+1}}{c_n}\leq \frac{(n+1)\log(n+1)}{n\log n}$ for all $n$; and for that it is enough to have
$$c_{n+1}\leq \frac{n+1}n\, c_n\, .$$
It is also required that 
$$\limsup c_n>0\qquad{\rm and}\qquad\sum_{n\geq 2} \frac{c_n}{n\log n}<\infty\, .$$
Set $n_1:=2$, and let $(n_k)_{k\geq 1}$ be a fast increasing sequence of integers, fast enough to have
$$ \frac{n_{k+1}}{n_k}\uparrow\infty\qquad{\rm and}\qquad\sum_{k=1}^\infty \frac{1}{\log(n_k)}<\infty\, .$$
Now, define a sequence $(c_n)$ as follows:
$$c_{n}=\frac{n}{n_{k+1}}\qquad{\rm if}\qquad n_k\leq n<n_{k+1}\cdot$$ 
Then $\limsup c_n\geq 1$ because $c_{n_{k+1}-1}=\frac{n_{k+1}-1}{n_{k+1}}\to 1$ as $k\to\infty$. Also, $c_{n+1}\leq \frac{n+1}nc_n$ for all $n$. Finally,
\begin{eqnarray}
\sum_{n=2}^\infty \frac{c_n}{n\log n}&=&\sum_{k=1}^\infty\frac1{n_{k+1}}\sum_{n_k\leq n<n_{k+1}}\frac{1}{\log n}\\
&\leq&\sum_{k=1}^\infty\frac1{n_{k+1}}\times \frac{n_{k+1}-n_k}{\log(n_k)}\\
&\leq&\sum_{k=1}^\infty \frac{1}{\log(n_k)}<\infty\, .
\end{eqnarray}
There is nothing special with $\log n$ here. The same proof shows that for any increasing function $\phi:\mathbb N\to\mathbb R^+$ tending to $\infty$, one can find a decreasing sequence $(a_n)$ such that $\sum a_n$ is convergent and $n\phi(n)\, a_n\not\to 0$.
