Is this easy to prove? $\forall N, k \gt 0$, $\pi(x) - \pi(N) \gt \frac{x-N}{2k}$ for all sufficiently large $x$.

Question

Let $\pi(x)$ be the prime counting function. Knowing that $\pi(x) \sim \dfrac{x}{\ln x}$. How could you prove that

$\pi(x) - \pi(N) \gt \dfrac{x - N}{2k}$

for all $x \geq $ some $X_0$?

I think it's false.

Answer 1

This is false for all $k,N$.

Since $\pi(x)\sim\frac{x}{\ln x}$, we have for all sufficiently large $x$ $$\pi(x)<\frac{2x}{\ln x}$$

Now it's enough to show that $\frac{2x}{\ln x}<\frac{x-N}{2k}+\pi(N)$ for any $N,k$ and $x$ large enough. This is true as RHS is linear function and LHS is sublinear.

Answer 2

The left-hand side grows slower than the right-hand side. So it will be false for $x$ large enough.