A sequence $\{a_n:n\in\mathbb{Z}_+\}\subset\{(0,\infty)$ is convex if $$ a_{n-1}+a_{n+1}-2a_n\geq0,\quad n\geq1 $$ This is equivalent to saying that $\{a_n-a_{n+1}:n\in\mathbb{Z}_+\}$ is a monotone non increasing sequence.
Convex sequences that converge to $0$ (i.e. $\lim_na_n=0$) are useful in determining integrability properties of trigonometric series. In this setting, it is obvious that $$ \sum_n(a_n-a_{n+1})=a_0 \tag{1}\label{one} $$ What I am trying to determine is why $\lim_nn(a_n-a_{n+1})=0$ holds.
I can see this is not too surprising for if $b_n=a_n-a_{n+1}$, then from the convergence of $\eqref{one}$ it follows that $b_n\geq0$ and so, $\sum_k 2^kb_k<\infty$ by a theorem of Cauchy. Hence $\lim_n nb_n=0$ holds along the subsequence $n_k=2^k$. A hint will be appreciated.
You may refer to this: Series converges implies $\lim{n a_n} = 0$
HINT
Let $b_n = a_n - a_{n+1}$.
$(b_n)$ is non-negative and non-increasing, and as you mentioned, $\sum_n b_n$ converges.
Try finding an upper bound of $n b_n$ that consists in the sum of $n$ terms of the sequence. Now show that this upper bound converges to $0$ as $n$ goes to infinity, using the fact that the series $\sum_n b_n$ converges.
SOLUTION
$0 \leq n b_n \leq \sum_{k=n}^{2n-1} b_k = \sum_{k=0}^{2n} b_k - \sum_{k=0}^{n} b_k \xrightarrow[n \to \infty]{} 0$