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

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

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)

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## marked as duplicate by Yiorgos S. Smyrlis, TMM, vonbrand, Sami Ben Romdhane, Andrew D. HwangMar 7 '14 at 20:41

– rlartiga Mar 7 '14 at 18:29
You've sort of done the two key things, but just not at the same time! Try fully factoring, rather than just going part way. – Hurkyl Mar 7 '14 at 18:29

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 $10001/10000$.

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Thanks! I should have gone into that direction more deeply but anyway this is a lesson that I should always remember: "Try all the pistes even if they seem irrelevant" – LoveFood Mar 7 '14 at 18:35

\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}

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Thank you man, but is there some trick that doesn't use the pi notation ($\prod$) like Gauss' trick to evaluate that product? I will be very thankful if you help me. – LoveFood Mar 10 '14 at 11:45
@LoveFood: don't fear the $\prod$ notation; it is very useful for writing products of a sequence in a compact way. I am not sure which trick of Gauss you mean. – robjohn Mar 10 '14 at 12:00

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

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

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