Have I proved $\sum_{i=1}^{\lg n} 2^{i-1} = \Theta(n\lg n)$? I have an exercise problem and don't know why its answer is like this.
$$ \sum_{i=1}^{\lg n} 2^{i-1} \in \Theta(2^{\lg n}) = \Theta(n).  $$

Regarding this equation, I think it would be, 
$$ \sum_{i=1}^{\lg n} 2^{i-1} \in \Theta(\lg n \cdot 2^{\lg n}) = \Theta(\lg n \cdot n),  $$
with the proof,
$$ \sum_{i=1}^{g(n)}f(i) \le \sum_{i=1}^{g(n)}f(g(n)) = g(n)\cdot f(g(n)).  $$

My thought with this question,  $ f(n) =  2^{i-1} $ and $ g(n)= \lg n $. Then result would be $ \lg n \cdot 2^{\lg n - 1} $.

Can anyone point out where I made a mistake? (I am not asking grading mine, but tell me where I mis-think)
Your argument shows that
$$
\sum_{i=1}^{\lg n} 2^{i-1} = O(n\lg n).
$$
It doesn't show the matching lower bound, which is actually false.