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Statement

For positive integers $a$ and $k$,

$$\frac{2^{(a+1)k-1}}{\sqrt{k}} \leq\binom{2^ak}{k}. \tag{8}$$

Ecklund prints Lemma 3 (p.268) as display (8) alone, with $n$ for the power parameter written $a$ here and no stated range; he says it is proved by induction on that parameter for all values of $k$. The range $a\geq1$ is needed, since at $a=0$ the left side exceeds $\binom kk=1$ once $k\geq2$. The theorem applies it with $a=2,3,4$.

Proof

First prove the case $a=1$. Set

$$A_k=\binom{2k}{k}\frac{\sqrt{k}}{4^k}.$$

Here $A_1=1/2$, and

$$\frac{A_{k+1}}{A_k} =\frac{2k+1}{2(k+1)}\sqrt{\frac{k+1}{k}}>1,$$

because the square of the last expression exceeds one by $1/(4k(k+1))$. Hence

$$\binom{2k}{k}\geq\frac{4^k}{2\sqrt{k}},$$

which is (8) for $a=1$.

For the induction step,

$$\frac{\binom{2^{a+1}k}{k}}{\binom{2^ak}{k}} =\prod_{j=0}^{k-1} \frac{2^{a+1}k-j}{2^ak-j} \geq2^k.$$

Multiplying the inductive bound by $2^k$ gives (8) with $a+1$ in place of $a$.

Verification record

Current review state. Accepted by independent mathematical review, retained as the full-proof review and its final receipt. Substantive changes to this reconstruction invalidate the affected scope until rechecked.

Scope and source version. The checked scope is equation (8), the central-binomial base case, and the induction on the power parameter for every positive integer $k$. Equation (8) is on printed p.268 / physical p.3 of Ecklund's Pacific Journal of Mathematics 29 (1969), 267--270 publisher PDF, identified on the source card.

Premises and limits. Ecklund prints the result and says only that induction proves it. The proof above is the compilation's expanded reconstruction, not a verbatim source proof, and it uses no external theorem. No gap remains inside the reconstructed induction at the accepted scope. No formal verification is recorded.