Rudy Koch

2022-01-24

Evaluate the infinite product
$\underset{n\to \mathrm{\infty }}{lim}\prod _{k=2}^{n}\left(1-\frac{1}{{k}^{2}}\right)$

becky4208fj

Expert

Note that
$1-\frac{1}{{k}^{2}}=\left(1-\frac{1}{k}\right)\left(1+\frac{1}{k}\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{{a}_{k}}{{a}_{k-1}}$
with ${a}_{k}=\frac{k+1}{k}$, hence this is a telescoping product, i.e.
$\prod _{k=2}^{n}\left(1-\frac{1}{{k}^{2}}\right)=\frac{{a}_{2}}{{a}_{1}}\frac{{a}_{3}}{{a}_{2}}\dots \frac{{a}_{n}}{{a}_{n-1}}=\frac{{a}_{n}}{{a}_{1}}=\frac{n+1}{2n}$

ocretz56

Expert

Let $g\left(k\right)=\frac{k-1}{k}$. Then this product is:
$\prod _{k=2}^{n}\frac{g\left(k\right)}{g\left(k+1\right)}$
which is a telescoping product, and thus equal to $\frac{g\left(2\right)}{g\left(n+1\right)}$

RizerMix

Expert

$\left(1-\frac{1}{\left(k-1{\right)}^{2}}\right)\left(1-\frac{1}{\left(k{\right)}^{2}}\right)\left(1-\frac{1}{\left(k+1{\right)}^{2}}\right)$ $=\frac{k\left(k-2\right)\left(k-1\right)\left(k+1\right)k\left(k+2\right)}{\left(k-1{\right)}^{2}{k}^{2}\left(k+1{\right)}^{2}}=\frac{\left(k-2\right)\left(k+2\right)}{\left(k-1\right)\left(k+1\right)}$ Observe that $\left(1-\frac{1}{\left(k{\right)}^{2}}\right)$ is cancelled out by the previous & the next term except for the extreme terms, the 1st & the last term leaving behind the 1st part of the 1st term $=\frac{2-1}{2}$ and the 2nd part of the last term $=\frac{n+1}{n}$