This question is possibly related to my previous question and it's answer.
Suppose $a_n>0\ \forall\ n\in\mathbb{N}\ $ and $\displaystyle\sum_{n=1}^{\infty} a_n = a.$
Does there exist a subsequence $(a_{k_n})_{n\in\mathbb{N}}$ of $(a_n)_{n\in\mathbb{N}}$ such that the sequence defined by:
$b_1 = a_{k_1}=a_1;\ $ and for each $n\in\mathbb{N},\ b_i = a_{k_n} + \frac{i-k_n}{k_{n+1}-k_n}\left( a_{k_{n+1}} - a_{k_n} \right)\ $ for all $i$ with $k_n < i \leq k_{n+1},$
has the properties that $(b_n)$ is convex and $\displaystyle\sum_{n=1}^{\infty} b_n = a?$
Note that without the restriction $a_n >0,\ $ the sequence $(a_n)= 1,0,0,0,0,0,0,0,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25},\ldots$ would be a counter-example.
There are many variations of this problem, but out of all of them, this one stuck.