Show that $∏_{k=1}^{∞}(u_{k}×4k^2)/(4k^2-1)<π/2$

Question

We know that $$∏_{k=1}^{∞}4k^2/(4k^2-1)=π/2$$

Let us consider a sequence $(u_{k})_{k≥1}$ such that $0≤u_{k}<1$ for all $k≥1$.

Show that $$∏_{k=1}^{∞}(u_{k}×4k^2)/(4k^2-1)<π/2$$

Answer 1

$\frac {(u_k)(4k^{2})} {4k^{2}-1} <\frac {4k^{2}} {4k^{2}-1}$ for all $k$. Just multiply these inequalities over $k \geq 2$ and note that we have strict inequality for $k=1$.

Question of convergence of the product: Note that $u_1u_2...u_n$ is decreasing and so it has a limit. Consider the cases when the limit is $0$ and the cases when the limit is positive to see that the intended inequality holds in all cases.

Answer 2

Every partial product satisfies $$ \prod_{k=1}^{N}\frac{u_{k} 4k^2}{4k^2-1} = (u_1 u_2 \cdots u_N)\prod_{k=1}^{N}\frac{ 4k^2}{4k^2-1} \le u_1 \prod_{k=1}^{N}\frac{ 4k^2}{4k^2-1} $$ so that $$ \prod_{k=1}^{\infty }\frac{u_{k} 4k^2}{4k^2-1} \le u_1 \prod_{k=1}^{\infty}\frac{ 4k^2}{4k^2-1} = u_1 \frac{\pi}{2} < \frac{\pi}{2} \, . $$