$M\in\mathbb{N}$ such that $\pi(n)\leq\frac{Mn}{\log n}$

Question

I know that $\pi(n)$ is approximately $\frac{n}{\log n}$. Is there any constant $M \in \mathbb{N}$ that satisfies:

$\forall n \in \mathbb{N}~.~\pi(n) \leq \frac{Mn}{\log n}$

I need the upper bound $M$ for computational analysis. Thanks!

Answer 1

For $n\geq55$, $\pi (n)<\frac{n}{\ln n-4}$ (Rosser, 1941), which in turn is bounded by $\frac{2n}{\ln n}$, for $x>2980$. So it suffices to find a bound for the finite set of cases $1\leq n\leq2980$. You can verify this fairly easily with a computer, so $M=2$ suffices. However, we have $\pi(7)>\frac{7}{\ln 7}$, so $M=2$ must be the smallest possible choice.

Answer 2

$$\frac{\pi(x)\log x}{x}$$ attains its maximum at $x = 113$. The value is $\approx 1.25506$. So if you need to take your $M$ as an integer, $2$ is the smallest admissible choice.