Prove: $4^{2k+1}>1 \cdot 3 \cdot \dots \cdot (2k+1)$ where $k$ is a positive integer

Question

Prove: $$4^{2k+1}>1 \cdot 3 \cdot \dots \cdot (2k+1)$$ where $k$ is a positive integer.

The difference is very large but I cannot find any way to prove it.

Answer 1

This isn't true. Indeed, the ratio $\frac{4^{2k+1}}{1\cdot 3\cdot...\cdot(2k+1)}$ tends to zero (while if your inequality was true, it would be bounded below by $1$).

To see this, let $a_k=\frac{4^{2k+1}}{1\cdot 3\cdot...\cdot(2k+1)}$. It's then easy to see that $\frac{a_{k+1}}{a_k}=\frac{16}{2k+3}$, which is less than $\frac{1}{2}$ for $k\geq 15$. So we have for $k\geq 15$:

$$a_k=a_{15}\cdot\frac{a_{16}}{a_{15}}\cdot\frac{a_{17}}{a_{16}}\cdot...\cdot\frac{a_k}{a_{k-1}}<a_{15}\cdot\frac{1}{2^{k-15}}$$

which clearly tends to zero.

Answer 2

The thing you wrote in the right hand side is the semifactorial $(2k+1)!!=\prod_{i\le 2k+1, i\equiv 2k+1p\mod{2}}i$. It turns our that $$ \forall i=1,\ldots,k, 2i+1>2i \implies (2k+1)!=(2k+1)!!(2k)!!<(2k+1)!!^2 $$ Therefore for all $k>64$ $$ (2k+1)!!>\sqrt{(2k+1)!}>\sqrt{63!\cdot 64^{2k+1-63}}>c\cdot 8^k $$ for a positive constant $c$. Therefore $(2k+1)!!>4^{2k+1}$ whenever $k$ is sufficiently large.

Answer 3

The statement is mistaken: the inequality should have been reversed. Let us prove then that $4^{2k+1} < 1 \cdot 3 \cdot \dots \cdot (2k+1)$ for $k \ge 19$ by induction.

Let us verify it for $k=19$: $4^{39} = 302231454903657293676544 < 319830986772877770815625 = 1 \cdot 3 \cdot \dots \cdot 37 \cdot 39$.

Assume the statement true for $k$ and let us prove it for $k+1$:

$$4^{2(k+1)+1} = 4^{2k+1} \cdot 16 < 1 \cdot 3 \cdot \dots \cdot (2k+1) \cdot 16 < 1 \cdot 3 \cdot \dots \cdot (2k+1) (2k+3)$$

because clearly $16 < 2k+3$ whenever $k \ge 19$.

(Note that the given inequality is not true for $0 \le k \le 18$.)