Suppose $x_n$ is a decreasing sequence of positive reals with $\sum x_n$ converges, must $(n\log n)x_n \to 0$ We are able to show, using the Cauchy criterion (using sum from $n$ to $2n$) that $nx_n \to 0$ Explicitly this is $0<nx_{2n}<\displaystyle \sum_{i=n}^{2n}x_n$ and the result follows from squeeze rule. My question is can we do better?
Intuitively, it feels like we should since $\frac{n\log n}{n^k} \to 0$ for $k>1$. Unfortunately, I am struggling to prove this, Any hints would be much appreciated.
One idea I had was to say that if $\displaystyle \sum_{i=N}^{n} x_i > \sum_{i=N}^{n} \frac{1}{n^k}$ for $k>1$ and for all sufficiently large $n$ then we could somehow compare it to the harmonic series. Could we not say the tail of the series could be made as large as we want this way and so divergence would be inevitable?
 A: Let $N_k = 2^{k^2}$ and define $(x_n)$ by $x_1 = 1$ and $x_n = \frac{1}{N_k \log N_k}$ if $N_{k-1} < n \leq N_k$ and $k \geq 1$. Then $(x_n)$ is positive and (weakly) decreasing. Also,
$$ \sum_{n=1}^{\infty} x_n
= 1 + \sum_{k=1}^{\infty} \frac{N_k - N_{k-1}}{N_k \log N_k}
\leq 1 + \sum_{k=1}^{\infty} \frac{1}{k^2\log 2}
< \infty. $$
On the other hand, $(n \log n) x_n = 1$ for infinitely many $n$, and so, it does not converge to $0$.
A: This holds when additionally $nx_n$ decreases to $0$. 
Indeed let us state the following lemma: if $(b_n)$ is a sequence that decreases to $0$, $\sum_n \frac{b_n}{n} <\infty \implies b_n=o\left( \frac 1{\log n}\right)$. 
To prove the lemma, note that by condensation, $\sum_n b_{2^n}$ converges, hence $b_{2^n}=o\left(\frac 1n\right)$ and $b_n=o\left( \frac 1{\log n}\right)$.
Applying the lemma with $b_n = nx_n$ yields the claim.
A: Hint: show that the series
$$ \sum_{n=2}^\infty \frac{1}{n (\ln n)(\ln \ln n)}$$
is divergent. From this it follows that $x_n$ must decrease faster, if $\sum_n x_n$ is to be convergent.
