Sum of powers of 2 from 1 to log(N). I came across the following sum: $\sum_{m=1}^{\log_2(N)} 2^{m}$. My intuition tells me that this should be bounded by 2N, but how would I prove this?

For any $p \in \mathbb{N}$ you have:
$$\sum_{m=1}^p 2^m = 2^{p+1}-2=2\cdot2^p - 2$$
So:
$$\sum_{m=1}^{\log_2(N)} 2^m = 2N-2$$ Which is indeed bounded by $2N$
