Prove that $(1-\frac{1}{2^2}\cdots 1-\frac{1}{9\,999^2})(1-\frac{1}{10\,000^2})=0.500\,05$

Question

Prove that $\displaystyle\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{9\,999^2}\right)\left(1-\frac{1}{10\,000^2}\right)=0.500\,05$

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Here are all my attempts to solve this problem:

So the first thing I thought about is to transform the expression of the form $$\left(1-\frac{1}{n^2}\right)$$

To the expression: $$\left(\frac{n^2-1}{n^2}\right)$$

But in evaluating the product $$\prod_{k=2}^n \frac{n^2-1}{n^2}$$ seems way complicated than what I'm capable of, I don't know how to evaluate products even if I know that notation.

The other idea is to transform again $$\left(1-\frac{1}{n^2}\right)\to \left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)$$

But this isn't helpful in any way.

My other attempts were to try to see what happens when I start evaluating this sum, and it turns out that a lot of things cancel out but again no result.

I'll be happy if someone could guide me to solve this problem. (I feel that there is some kind of symmetry that I should remark, a symmetry that would allow me to cancel things out and to have my final)

Answer 1

Note that you can factor $n^2 - 1 = (n - 1)(n + 1)$ to find something like

\begin{align*} \left(1 - \frac 1 {2^2}\right)&\left(1 - \frac 1 {3^2}\right) \cdots \left(1 - \frac 1 {10000^2}\right) \\ & =\left(\frac 1 2 \cdot\frac 3 2\right) \left(\frac 2 3 \cdot\frac 4 3\right) \left(\frac 3 4 \cdot \frac 5 4\right) \left(\frac 4 5 \cdot \frac 6 5 \right)\cdots \left(\frac{10000 - 1}{10000} \cdot \frac{10000+1}{10000}\right) \end{align*} Convince yourself that everything cancels (i.e. the product telescopes), except for a factor of $1/2$, and the final factor of $10001/10000$.

Answer 2

$$ \begin{align} \prod_{k=2}^n\left(1-\frac1{k^2}\right) &=\prod_{k=2}^n\frac{(k-1)(k+1)}{k^2}\\ &=\prod_{k=2}^n\frac{k-1}{k}\hspace{1.2cm}\prod_{k=3}^{n+1}\frac{k}{k-1}\\ &=\frac12\left(\prod_{k=3}^n\frac{k-1}{k}\right)\frac{n+1}{n}\left(\prod_{k=3}^n\frac{k}{k-1}\right)\\ &=\frac12\frac{n+1}{n}\\ &=\frac{n+1}{2n} \end{align} $$

Answer 3

$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$

After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$

$$\frac{n+1}{2}\cdot\frac{1}{n}$$

Substitute the value of n=10000

and get the result