How to solve this summation of ceiling function in BUILD-MAX-HEAP algorithm I am stuck on solving this problem and cannot understand how is the ceiling function omitted or solved.
Please help.

The equation:

$\sum_{h=0}^{\lfloor\lg n\rfloor} \lceil\frac{n}{2^{h+1}}\rceil O(h) $. 

But this transformed to:

$O\left(n \sum_{h=0}^{\lfloor\lg n\rfloor} \frac{h}{2^h}\right)$

My question concerns on omitting the ceiling function. I'm not clear on whether it was omitted or solved.

This has been taken from CLRS Section $6.3$ Building a Heap.

Please help me out.
Thank you.
$$
\begin{align*}
\sum_{h=0}^{\lfloor\lg n\rfloor} \left\lceil\frac{n}{2^{h+1}}\right\rceil O(h) 
&amp;&lt;
\sum_{h=0}^{\lfloor\lg n\rfloor} \left(1+ \frac{n}{2^{h+1}}\right) O(h)  \\
&amp;&lt;
\sum_{h=0}^{\lfloor\lg n\rfloor} \left(\frac{n}{2^h} + \frac{n}{2^h}\right) O(h) 
=
2n \sum_{h=0}^{\lfloor\lg n\rfloor} \frac{ O(h) }{2^h}
= O\left(n \sum_{h=0}^{\lfloor\lg n\rfloor} \frac{ h }{2^h}\right).
\end{align*}
$$.