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
$$\ln n \sim \sum_{k=1}^{n}\frac{1}{k}, n\to \infty.$$

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?
