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$${\frac{\pi}{2} = \lim_{l \to \infty} \prod_{j = 1}^{l + 1} \frac{(2j)(2j)}{(2j - 1)(2j+1)}}$$

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 $l$?

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    $\begingroup$ Yes. It is rather odd to have the $l+1$ rather than $l$. $\endgroup$ – copper.hat Jan 16 '16 at 4:46
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    $\begingroup$ Maybe there's some reason the formula was written this particular way in the context where this equation originally appeared. Where was that? $\endgroup$ – David K Jan 16 '16 at 4:49
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    $\begingroup$ this is from rochester.edu/newscenter/… $\endgroup$ – jessie Jan 16 '16 at 4:58
  • $\begingroup$ This is equivalent to the Wallis product for pi. Wallis was Newton's predecessor in the "Lucas chair" so he was (initially) pre-calculus. I dk how Wallis got his product, or when. $\endgroup$ – DanielWainfleet Jan 16 '16 at 8:42
  • $\begingroup$ As for where this expression comes from, use $x=\frac{\pi}{2}$ in $\frac{x}{\sin x}=\prod_{j=1}^\infty\frac{1}{1-x^2/(\pi^2j^2)}$. $\endgroup$ – J.G. Jun 20 '19 at 9:25
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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, $$ \begin{align} \lim_{\ell\to\infty}\prod_{j=1}^\ell\frac{2j}{2j-1}\frac{2j}{2j+1} &=\frac12\frac{\Gamma\left(\frac12\right)^2}{\Gamma(1)^2}\cdot1\\ &=\frac\pi2 \end{align} $$

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  • $\begingroup$ You should say $l\to \infty$ if and only if $l-1\to \infty$ in order to conclude $\lim_{l\to\infty}=\lim_{l-1\to\infty}$. $\endgroup$ – user236182 Jan 16 '16 at 7:08
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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

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    $\begingroup$ Thanks. It now more makes sense $\endgroup$ – jessie Jan 16 '16 at 5:01
  • $\begingroup$ Regarding infinite products: Sometimes the $\prod_{i=1}^\infty a_i$ is said not to converge even if $\lim_{n\to\infty}\prod_{i=1}^n a_i$ exists! $\endgroup$ – Hagen von Eitzen Jan 16 '16 at 10:09

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