Lower bound of a summation with an exponential
There is a general formula for this sum:
$$
\sum_{h=0}^m h2^h = \sum_{h=1}^m \sum_{k=1}^h 2^h = \sum_{k=1}^m \sum_{h=k}^m 2^h = \sum_{k=1}^m (2^{m+1}-2^k) = m2^{m+1} - (2^{m+1}-2).
$$
Overall, we get
$$
\sum_{h=0}^m h2^h = (m-1)2^{m+1} + 2.
$$
When $m = \lg n$, this works out to be $2n\lg n - 2n + 2$.