Convex-like property of a sequence Setting
Suppose we have an arbitrary sequence of natural numbers $a_1,\dots,a_n,\dots$ and two strictly positive numbers $h,K > 0$. We consider the sequence $(b_n)_n$ defined by:
$$
b_n := \frac{K+h\sum_{i=2}^n (i-1)a_i}{\sum_{i=1}^n a_i},
$$
if I wrote it correctly we should have $b_1 = K/a_1$, $b_2 = (K+ha_2)/(a_1+a_2)$, $b_3 = (K+ha_2+2ha_3)/(a_1+a_2+a_3)$ and so on.
Question
It is possible that $b_n < b_{n+1}$ and $b_n > b_{n+1}$,  but I would like to show that if $b_n \leq b_{n+1}$ then also $b_{n+1} \leq b_{n+2}$ and so on. I guess this would follow if we show that $b: \mathbb{N} \rightarrow \mathbb{R}: n \mapsto b_n$ is convex.
 A: First of all, we reformulate the setting as follows. Put $s_n=\sum_{i=1}^n a_i$. Then $\{a_n\}$ is a sequence of natural numbers iff 
$\{s_n\}$ is a increasing sequence of natural numbers. Next, 
$$b_n := \frac{K+h(ns_n-\sum_{i=1}^n s_i)}{s_n}=hn+\frac{K-h\sum_{i=1}^n s_i}{s_n}.$$
Now we can see that the sequence $\{b_n\}$ has no convexity even in a weak form $b_{n+1}\le\frac {b_n+b_{n+2}}2$. Indeed, put $K=1$, $h\simeq 0$, $s_1=3$, $s_2=4$, and $s_3=7$. Then 
$$b_2\simeq\frac 14=\frac 5{20}>\frac 5{21}=\frac 12\left( \frac 13+\frac 17\right)\simeq \frac {b_1+b_3}2.$$ 
Nevertheless, your conjecture is still true. In order to see that we reformulate condition $b_n\le b_{n+1}$ by the following equivalent transformations. 
$b_n\le b_{n+1}$
$hn+\frac{K-h\sum_{i=1}^n s_i}{s_n}\le h(n+1)+\frac{K-h\sum_{i=1}^{n+1} s_i}{s_{n+1}}$
$s_{n+1}\left(K-h\sum_{i=1}^n s_i\right)\le hs_ns_{n+1}+s_n\left(K-h\sum_{i=1}^{n+1} s_i\right)$
$K\le\frac h{s_{n+1}-s_n}\left(s_{n+1}\sum_{i=1}^n s_i+s_ns_{n+1}-s_n\sum_{i=1}^{n+1} s_i \right)$
$K\le\frac h{s_{n+1}-s_n}\left(s_{n+1}\sum_{i=1}^n s_i-s_n\sum_{i=1}^n s_i \right)$
$K\le h\sum_{i=1}^n s_i$.
Thus $b_n\le b_{n+1} \Leftrightarrow K\le h\sum_{i=1}^n s_i\Rightarrow  K\le h\sum_{i=1}^{n+1} s_i \Leftrightarrow
b_{n+1}\le b_{n+2}$, because the sequence $\{s_n\}$ consists of non-negative numbers (and we even don’t need that it is increasing).
