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?

$$\sum_{n=0}^d x^n= \frac{x^{d+1}-1}{x-1}$$
So we have:
$$\sum_{m=1}^{\log_2 N} 2^m=\sum_{m=0}^{\log_2 N} 2^m-2^0=\sum_{m=0}^{ \log_2 N } 2^m-1 =  \frac{2^{\lfloor \log_2N \rfloor +1}-1}{2-1}-1 \leq  \frac{N \cdot 2-1}{1}-1 \\=N \cdot 2-2 $$
