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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.

Mittens
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  • Every $a_n - a_{n+1}$ is positive. What happens if frequently in $n$, you have $n(a_n - a_{n+1}) >\varepsilon>0$? – Exodd Jun 24 '20 at 17:21
  • Once you have that, it is easy to conclude – Exodd Jun 24 '20 at 17:30
  • If for every $\varepsilon>0$, you know that $ n(a_n - a_{n+1}) <\varepsilon$ definitively in $n$, then by definition of limit $n(a_n - a_{n+1}) \to 0$ – Exodd Jun 24 '20 at 17:40

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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

$0 \leq n b_n \leq \sum_{k=n}^{2n-1} b_k = \sum_{k=0}^{2n} b_k - \sum_{k=0}^{n} b_k \xrightarrow[n \to \infty]{} 0$

DodoDuQuercy
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    @DooDuQuercy: As I suspected, it was embarrassingly obvious, but I was not able to se it. Thanks for the reference. In particular the solution that uses Cauchy's domination. – Mittens Jun 24 '20 at 17:57