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Source and scope. This elementary expansion supplies the comparison with $2^N/\sqrt N$ used in Erdős (1945), Theorem 2, printed p. 899 (published scan). It is not a separately numbered lemma in the paper.

Statement. For every integer $N\ge1$,

$$\frac{2^N}{2\sqrt N} \le B_N \le \frac{\sqrt2\,2^N}{\sqrt N}, \qquad B_N=\binom N{\lfloor N/2\rfloor}.$$

Proof. Set $q_m=\binom{2m}m/4^m$, so $q_0=1$ and

$$\frac{q_m}{q_{m-1}}=\frac{2m-1}{2m} \qquad(m\ge1).$$

We first show

$$\frac1{2\sqrt m}\le q_m\le\frac1{\sqrt{m+1}} \qquad(m\ge1).$$

For the lower bound, $q_1=1/2$. For $m\ge2$,

$$\left(\frac{2m-1}{2m}\right)^2 \ge\frac{m-1}{m}.$$

The recurrence therefore carries $q_{m-1}\ge1/(2\sqrt{m-1})$ to $q_m\ge1/(2\sqrt m)$. For the upper bound, start with $q_0=1$ and use

$$\left(\frac{2m-1}{2m}\right)^2 \le\frac m{m+1} \qquad(m\ge1).$$

After clearing the positive denominators, this inequality is $1\le3m$. It carries $q_{m-1}\le1/\sqrt m$ to $q_m\le1/\sqrt{m+1}$.

For even $N=2m$, $B_N/2^N=q_m$. For odd $N=2m-1$, $\binom{2m}m=2\binom{2m-1}{m-1}$ gives the same identity. Thus for every $N\ge1$,

$$\frac{B_N}{2^N}=q_{\lceil N/2\rceil}.$$

Since $\lceil N/2\rceil\le N$ and $\lceil N/2\rceil+1\ge N/2$, the preceding two bounds imply the statement. $\square$

Use. The constants suffice for Theorem 2 and the assertion that its order in $N$ cannot be improved at fixed radius one. No optimal constant or Stirling estimate is asserted.