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Several proofs pf Chebyshev's estimates for $\pi(n)$ include the following as a preliminary step (eg pdf):

$$\frac{2^{2n}}{2n} \leq {2n \choose n}$$

I can't see why this is true.

(I would appreciate answers accessible to readers without university training in mathematics.)

Bill Dubuque
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Penelope
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2 Answers2

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The $2n$ numbers $$\dbinom{2n}{1}, \dbinom{2n}{2}, \dbinom{2n}{3}, \ldots, \dbinom{2n}{n} \ldots,\dbinom{2n}{2n-2}, \dbinom{2n}{2n-1}, \dbinom{2n}{2n}+\dbinom{2n}{0} = 2$$ add up to $2^{2n}$, and $\dbinom{2n}{n}$ is the largest of these numbers. Since the largest number in a set is greater than or equal to the average of the numbers in the set, we have $\dbinom{2n}{n} \ge \dfrac{2^{2n}}{2n}$

JimmyK4542
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  • This explanation is very elegant! – Penelope Apr 04 '21 at 23:28
  • I just spotted something that I don't understand. There are 2n+1 terms, not 2n. That means the inequality is $\dbinom{2n}{n} \ge \dfrac{2^{2n}}{2n+1}$ ? – Penelope Apr 05 '21 at 03:40
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    The list I made has $2n$ elements. $2n-1$ of these elements are in the form $\binom{2n}{k}$ for $k = 1,\ldots,2k-1$. The last one is $\binom{2n}{2n}+\binom{2n}{0} = 2$. – JimmyK4542 Apr 05 '21 at 03:43
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Proceed by induction (I think induction is introduced in high school). It is clear for $ n = 1$.

Suppose now that $2^{2n}/2n \leq \binom{2n}{n}$. We show it holds for $n+1$

$\binom{2n+2}{n+1} = \frac{2n+2!}{n+1!n+1!} = \frac{2n+2 \cdot 2n+1}{n+1 \cdot n+1}\frac{2n!}{n!n!} \geq \frac{2\cdot2n+1}{n+1}\frac{2^{2n}}{2n} $ (by induction hypothesis) $ \geq \frac{2 \cdot 2n \cdot 2^{2n}}{(n+1)2n}$ (as $2n+1 \geq 2n$ of course) $= \frac{2^{2n+2}}{2(n+1)}$

and thus by induction we're done.

Vercingetorix
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