Prove or disprove: $\lim_{n\to\infty}(a_n n\ln n)=0$, where $a_n>0$, $a_{n+2}-a_{n+1}\geq a_{n+1}-a_n$, and $\sum_{k=1}^{n}a_n$ is bounded 
Suppose, for all $n \in \mathbb{N^+}$, it holds that

*

*(1) $a_n>0$;

*(2) $a_{n+2}-a_{n+1}\geq a_{n+1}-a_n$;

*(3) $\sum_{k=1}^{n}a_n$ is bounded.


Prove or disprove
$$\lim\limits_{n \to \infty} (a_n\cdot n\cdot\ln n )=0$$
First, since (1) and (3), we can obtain $\sum_{n=1}^{\infty}a_n$ is convergent,hence $\lim\limits_{n \to \infty}{a_n}=0.$ Now,notice that $\sum_{k=1}^{n-1}(a_{k+1}-a_{k})=a_n-a_1$, and let $n \to \infty$, we can obtain $\sum_{n=1}^{\infty}(a_{n+1}-a_n)=-a_1$.But how to go on with these? Maybe we may consider
$$\ln n \sim \sum_{k=1}^{n}\frac{1}{k}, n\to \infty.$$
 A: Only considerations.
$\displaystyle b_n := \frac{1}{n\ln(n+1)\ln\ln(n+2)}$
$(1)~~b_n>0~~$ o.k.
$(2)~~b_n-b_{n+1} \geq b_{n+1}-b_{n+2}~~$ o.k.
$(3)~~\sum\limits_{k=1}^n b_n~~$ is not bounded 
But we get: $~~\lim\limits_{n\to\infty}b_n n\ln n\to 0$
This means: If $~\sum\limits_{k=1}^n a_n~$ is bounded then it exists $~N~$ so that for all $~n\geq N~$ we have $~0<a_n<b_n~$ . Therefore we get $~~\lim\limits_{n\to\infty}a_n n\ln n\to 0 ~$ .
Is something missing in these considerations?
A: A few observations:
Since $\exists M>0, \forall n, 0<\sum_{k=1}^n a_k\leq M$, each $a_k$ is in $[0,M]$.
Since $a_{n+1}-a_n$ is non-decreasing and bounded, it converges to some $\ell$. If $\ell \neq0$, then by Cesaro, $a_n\sim \ell n$ and $a_n$ is unbounded, a contradiction. Thus $\lim_n (a_{n+1}-a_n)=0$.  
$a_{n+1}-a_n$ is non-decreasing and goes to $0$, so it's $\leq 0$, hence $a_n$ is non-increasing, so it converges to some $\ell'$. Since $\sum_{k=1}^n a_k$ is bounded, we must have $\ell'=0$, hence $\lim_n a_n=0$.
So 


*

*$a_n$ decreases to $0$

*$a_n-a_{n+1}$ decreases to $0$

*$\sum_n a_n$ converges


Since $a_n$ is non-increasing, it suffices to prove the claim for $n=2^k$, i.e. $2^ka_{2^k}=o\left( \frac 1 k\right)$. Under the additional assumption that $na_n$ is non-increasing, this follows from Cauchy's condensation test.
It may be interesting to consider the continuous analog: if $f:\mathbb R_+ \to \mathbb R_+$ is decreasing, convex and $\int_0^\infty f(t)dt<\infty$, do we have $$f(x)=o\left(\frac{1}{x\ln x} \right)$$
