Why does this equation work?

Question

let $ P(x) := \sum_{p \leq x} Log [p]$, then we have

• $P(2^{k+1}) = \sum_{i=0}^k ( P(2^{i+1}) - P(2^i)) < 2 \cdot Log[2] \cdot (1 + 2 + 4 +... + 2^k) \leq 4 \cdot Log[2] \cdot 2^k$.

Why does this equation work?

The first step is clear for me, since we have a telescope sum. However, how can I conclude to the first $<$ and the $\leq$ ?

Answer 1

The problem is solved now, the solution is pretty simple, since

• $P(2n) - P(n) < 2\cdot Log[2] \cdot n$,

which can be shown fast, we can conclude the above inequality by setting

• $2n = 2^{i+1}$ and $n = 2^{i+1} / 2 = 2^i$ and then summing up.

Thanks for your help.