Proving that:
$$ \sum_{j=1}^{\log_2m}m/2^j \geq m-1 $$
I tried to use geometric serie and use the formula:
$$ S = a_1 \frac{q^n-1}{q-1} $$
Changed to:
$$ m\sum_{j=1}^{\log_2m}1/2^j $$
For
$$ a_1 = 1/2, q = 1/2, n = \log_2m $$
Getting:
$$ m\sum_{j=1}^{\log_2m}1/2^j = m(1-0.5^{\log_2m}) $$
But I don't see how it satisfies:
$$ m(1-0.5^{\log_2m}) \geq m-1 $$
In fact $m(1-0.5^{log_2 m}) = m-1$. Since $0.5^{log_2 m}=2^{-log_2 m}=1/m$.