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I may not use the Prime Number Theorem (and even if I could, it would only show the desired claim for sufficiently-large $n$).

I have tried induction on $n^{\pi\left(2n\right)-\pi\left(n\right)}<2^{2n}$. For the induction step we have: $$n^{\pi\left(2n+2\right)-\pi\left(n+1\right)} \le n^{\pi\left(2n+2\right)-\pi\left(n\right)} \le n^{\pi\left(2n\right)+1-\pi\left(n\right)} <2^{2n}n$$

Which isn't less than $2^{2n+2}$.

Any advice is very welcome.

Adar Hefer
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    you can prove it by PNT for all number with few cases to be checked and not for sufficiently large $n$, but i am assuming you want induction way to prove the above. – Ahmad Aug 26 '17 at 11:25
  • It's surely a legitimate solution if I can pinpoint the cases to be manually checked, but how would I go about doing that? (I'm assuming this isn't the "intended" solution for this problem but I'm interested nonetheless) – Adar Hefer Aug 28 '17 at 10:54

2 Answers2

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There is a classical approach without much computation.

For any prime $p$, we may estimate its exponent in $\binom{2n}{n}$ as follows $$\nu_p\biggl(\binom{2n}{n}\biggr)=\sum_{k=1}^{\lfloor\log_p2n\rfloor}\biggl(\Bigl\lfloor\frac{2n}{p^k}\Bigr\rfloor-2\Bigl\lfloor\frac{n}{p^k}\Bigr\rfloor\biggr)\leqslant\sum_{k=1}^{\lfloor\log_p2n\rfloor}1\leqslant\log_p2n.$$ So we have $$\binom{2n}{n}=\prod_{\text{prime}~p\\p<2n}p^{\nu_p\bigl(\binom{2n}{n}\bigr)}\leqslant\prod_{\text{prime}~p\\p<2n}p^{\log_p2n}=\prod_{\text{prime}~p\\p<2n}2n=(2n)^{\pi(2n)}.$$ For $n>1$, we also have $$\binom{2n}{n}=\prod_{k=1}^n\frac{n+k}{k}>\prod_{k=1}^n2=2^n.$$ This proved the second inequality.

For the first one, just note that $\binom{2n}{n}$ is divisible by any prime $p$ in the region $(n,2n)$, therefore $$\binom{2n}{n}\geqslant\prod_{\text{prime}~p\\n<p<2n}p\geqslant\prod_{\text{prime}~p\\n<p<2n}n=n^{\pi(2n)-\pi(n)}.$$ Then $\binom{2n}{n}<\sum_{k=0}^{2n}\binom{2n}{k}=2^{2n}$ gives the disired result.

Mercury
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Using this result (and some discussions here)

$$\frac{x}{\ln{x}}<\pi(x)< 1.25506\frac{x}{\ln{x}} \text{, for } x \geq 17 \tag{1}$$

The second one $2^n \leq (2n)^{\pi(2n)}$ is a direct result of $\pi(x)>\frac{x}{\ln{x}}$ or $$\pi(2n)>\frac{2n}{\ln{(2n)}}>\frac{n\ln{2}}{\ln{(2n)}} \Rightarrow \pi(2n) \cdot \ln{(2n)} > n \cdot \ln{2} \Rightarrow \\ \ln{(2n)^{\pi(2n)}}>\ln{2^n} \Rightarrow (2n)^{\pi(2n)} > 2^n$$

The first one $$n^{\pi(2n)-\pi(n)}< 2^{2n} \Leftrightarrow \left(\pi(2n)-\pi(n) \right) \ln{n} < 2n \ln{2} \tag{2}$$ but $$1.25506 < 2\ln{2}$$ and it is enough to prove $$\color{red}{\pi(2n)-\pi(n) < \pi(n)} < 1.25506\frac{n}{\ln{n}} < 2 \ln{2}\frac{n}{\ln{n}}$$

which is covered here.

rtybase
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