We have $$π/2=∏_{k=1}^{∞}(2k)²/((2k-1)(2k+1))$$
Show that $$∏_{k=1}^{n}(2k)²/((2k-1)(2k+1))<π/2$$ for all $n≥1$.
I have remarked that $(2k)²/((2k-1)(2k+1))>1$ for all $k$
We have $$π/2=∏_{k=1}^{∞}(2k)²/((2k-1)(2k+1))$$
Show that $$∏_{k=1}^{n}(2k)²/((2k-1)(2k+1))<π/2$$ for all $n≥1$.
I have remarked that $(2k)²/((2k-1)(2k+1))>1$ for all $k$
Hint
$$a_n\triangleq \prod_{k=1}^{n}{(2k)^2\over (2k-1)(2k+1)}$$therefore $$a_{n+1}>a_n>1$$