We prove $\pi(x) < 2\frac{x}{\log x}$ for sufficiently large $x$.

Key lemma: For the central binomial coefficient: $\binom{2n}{n} < \frac{4^n}{\sqrt{n}}$

Consider the prime factorization of $\binom{2n}{n}$. For prime $p$, the exponent of $p$ in $n!$ is: $\sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor$

The exponent of $p$ in $\binom{2n}{n} = \frac{(2n)!}{n!^2}$ is: $\sum_{k=1}^\infty \left( \left\lfloor \frac{2n}{p^k} \right\rfloor - 2\left\lfloor \frac{n}{p^k} \right\rfloor \right)$

Each term is 0 or 1, and only terms with $p^k \leq 2n$ contribute.

Therefore: $\binom{2n}{n} = \prod_{p \leq 2n} p^{e_p}$ where $e_p \leq \log_{p}(2n)$.

Taking logarithms: $\log \binom{2n}{n} = \sum_{p \leq 2n} e_p \log p \leq \sum_{p \leq 2n} \log p \cdot \log_{p}(2n)$

Since $\log_{p}(2n) = \frac{\log(2n)}{\log p}$: $\sum_{p \leq 2n} \log p = \theta(2n)$

From the lemma: $\log \binom{2n}{n} < n \log 4 - \frac{1}{2}\log n$.

Combining: $\theta(2n) < n \log 4$.

Using $\theta(x) \sim \pi(x) \log x$ (which can be shown separately), we get: $\pi(x) \log x \lessapprox \theta(x) < \frac{x \log 4}{2}$

Therefore: $\pi(x) < \frac{x \log 4}{2\log x} \approx 0.69 \frac{x}{\log x}$ for large $x$.

More careful analysis gives the bound $\pi(x) < 1.11\frac{x}{\log x}$.

We prove that for $n \geq 1$, there exists a prime $p$ with $n < p < 2n$.

Proof using binomial coefficients:

Consider $\binom{2n}{n}$. We analyze its prime factorization carefully.

Claim 1: Every prime $p$ with $\frac{2n}{3} < p \leq n$ appears exactly once in $\binom{2n}{n}$.

Proof: For such $p$, we have $p \leq n < 2p$, so $\lfloor \frac{2n}{p} \rfloor = 1$ and $\lfloor \frac{n}{p} \rfloor = 0$.

Thus, the exponent is $1 - 2 \cdot 0 = 1$.

Claim 2: Every prime $p > 2n$ does not divide $\binom{2n}{n}$.

Claim 3: For $\sqrt{2n} < p \leq \frac{2n}{3}$, the exponent of $p$ is at most 1.

Proof: If $p^2 > 2n$, then at most one power of $p$ appears.

Lower bound: From the binomial theorem: $4^n = (1 + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} > \binom{2n}{n}$

Upper bound: Combining all primes: $\binom{2n}{n} \leq \prod_{p \leq \sqrt{2n}} (2n)^{\log(2n)/\log p} \cdot \prod_{\sqrt{2n} < p \leq 2n} p$

The first product is bounded by $(2n)^{\sqrt{2n}}$ (summing over $\pi(\sqrt{2n})$ terms).

If there were no primes in $(n, 2n]$, the second product would cover only primes up to $n$: $\binom{2n}{n} \leq (2n)^{\sqrt{2n}} \cdot \prod_{p \leq n} p < (2n)^{\sqrt{2n}} \cdot 4^n$

For large $n$, this contradicts $\binom{2n}{n} \geq \frac{4^n}{2n}$ (from Stirling).

Therefore, there must be a prime in $(n, 2n]$.

The analytic proof of the PNT uses the Riemann zeta function: $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_p \frac{1}{1 - p^{-s}}$

Key steps:

1. Show $\zeta(s)$ has an analytic continuation to $\Re(s) > 0$ except for a simple pole at $s = 1$
2. Prove $\zeta(s) \neq 0$ on the line $\Re(s) = 1$
3. Use contour integration to relate $\psi(x)$ to $\zeta(s)$: $\psi(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} -\frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s} ds$
4. Shift the contour left; the residue at $s = 1$ gives the main term $x$
5. Control error terms using bounds on $\zeta(s)$

The result: $\psi(x) \sim x$, which implies $\pi(x) \sim \frac{x}{\log x}$.

The proof requires deep results from complex analysis, including properties of the logarithmic derivative of $\zeta(s)$ and careful analysis near the critical line.