Can anyone explain this equation (about $\frac\pi2$ ) 
$${\frac{\pi}{2} = \lim_{l \to \infty} \prod_{j = 1}^{l + 1} \frac{(2j)(2j)}{(2j - 1)(2j+1)}}$$

Hi all.
My first impression of this equation is naive curiosity why "limit" is required. 
Can I just drop the limit sign and replace $l+1$ by $\infty$? 
Or If "limit" can not be omitted, why would we multiply all the terms upto $l+1$? 
Does it change anything if I replace $l+1$ by $l$? 
 A: These are all good questions. Some comments:
(1) Products with $\infty$ as the upper limit are not really products. You cannot multiply an infinite number of factors. Multiplication is defined as a binary operation, so you can only multiply two factors at a time. Repeating this many times allows a finite product; you can then try to take a limit to pass to the infinite case, but don't make the mistake of thinking that you are performing an infinite number of operations. Same thing with infinite sums. In other words, an "infinite product" is defined to be the limit of a sequence of partial products (and similarly for "infinite sums").
(2) $\ell +1$ is not significant, could use anything that grows without bound as $\ell\to\infty$
(3) No
A: 
Can I just drop the limit sign and replace $l+1$ by $\infty$?

These mean the same thing:
$$
\prod_{j=1}^\infty\frac{2j}{2j-1}\frac{2j}{2j+1}
\stackrel{\text{def}}{\equiv}\lim_{\ell\to\infty}\prod_{j=1}^\ell\frac{2j}{2j-1}\frac{2j}{2j+1}
$$


Does it change anything if I replace $l+1$ by $l$ ?

Since $\ell\to\infty\iff\ell-1\to\infty$, we have
$$
\begin{align}
\lim_{\ell\to\infty}\prod_{j=1}^\ell\frac{2j}{2j-1}\frac{2j}{2j+1}
&=\lim_{\ell-1\to\infty}\prod_{j=1}^\ell\frac{2j}{2j-1}\frac{2j}{2j+1}\\
&\equiv\lim_{\ell\to\infty}\prod_{j=1}^{\ell+1}\frac{2j}{2j-1}\frac{2j}{2j+1}
\end{align}
$$

One way to evaluate the infinite product
$$
\begin{align}
\prod_{j=1}^{\ell}\frac{2j}{2j-1}\frac{2j}{2j+1}
&=\frac{2^{2\ell}\ell!^2}{(2\ell)!}\frac{2^{2\ell+1}\ell!(\ell+1)!}{(2\ell+2)!}\\
&=\frac{2^{4\ell}\ell!^4}{(2\ell)!^2(2\ell+1)}\\
&=\left(\frac{4^\ell}{\binom{2\ell}{\ell}}\right)^2\frac1{2\ell+1}
\end{align}
$$
Using inequality $(9)$ from this answer, we get
$$
\frac{\pi\left(\ell+\frac14\right)}{2\ell+1}
\le\prod_{j=1}^{\ell}\frac{2j}{2j-1}\frac{2j}{2j+1}
\le\frac{\pi\left(\ell+\frac13\right)}{2\ell+1}
$$
Using the Squeeze Theorem, we get
$$
\lim_{\ell\to\infty}\prod_{j=1}^{\ell}\frac{2j}{2j-1}\frac{2j}{2j+1}=\frac\pi2
$$

Another way to evaluate the infinite product
$$
\begin{align}
\prod_{j=1}^\ell\frac{2j}{2j-1}\frac{2j}{2j+1}
&=\prod_{j=1}^\ell\frac{j}{j-\frac12}\frac{j}{j+\frac12}\\
&=\frac{\Gamma(\ell+1)/\Gamma(1)}{\Gamma\left(\ell+\frac12\right)/\Gamma\left(\frac12\right)}\frac{\Gamma(\ell+1)/\Gamma(1)}{\Gamma\left(\ell+\frac32\right)/\Gamma\left(\frac32\right)}\\
&=\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac32\right)}{\Gamma(1)^2}\frac{\Gamma(\ell+1)\Gamma(\ell+1)}{\Gamma\left(\ell+\frac12\right)\Gamma\left(\ell+\frac32\right)}\\
&=\frac12\frac{\Gamma\left(\frac12\right)^2}{\Gamma(1)^2}\frac{\Gamma(\ell+1)^2}{\Gamma\left(\ell+\frac12\right)^2}\frac1{\ell+\frac12}\\
\end{align}
$$
By Gautschi's Inequality,
$$
\frac\ell{\ell+\frac12}\le\frac{\Gamma(\ell+1)}{\Gamma\left(\ell+\frac12\right)}\frac1{\ell+\frac12}\le\frac{\ell+1}{\ell+\frac12}
$$
By the Squeeze Theorem,
$$
\begin{align}
\lim_{\ell\to\infty}\prod_{j=1}^\ell\frac{2j}{2j-1}\frac{2j}{2j+1}
&=\frac12\frac{\Gamma\left(\frac12\right)^2}{\Gamma(1)^2}\cdot1\\
&=\frac\pi2
\end{align}
$$
