# Evaluating the infinite product $\prod_{k=2}^{\infty }\left ( 1-\frac{1}{k^2} \right ).$

Evaluate

$$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$

I can't see anything in this limit , so help me please.

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Note that $$1-\frac1{k^2}=\left(1-\frac1k\right)\left(1+\frac1k\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{a_k}{a_{k-1}}$$ with $a_k= \frac{k+1}k$, hence this is a telescoping product, i.e. $$\prod_{k=2}^n\left(1-\frac1{k^2}\right)=\frac{a_2}{a_1}\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}=\frac{a_n}{a_1}=\frac{n+1}{2n}.$$

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I think that is \frac{a_n}{a_1} –  Sherloek holmes Dec 26 '12 at 19:24

$$1-\frac{1}{k^2} = \frac{(k-1)(k+1)}{k^2}$$ $$\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ) = \underbrace{\frac{2-1}{2}}_{\text{from first}} \times \underbrace{\frac{(n+1)}{n}}_{\text{from last}}$$

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Hint: The "typical" term is $\dfrac{k-1}{k}\dfrac{k+1}{k}$. Express the first few terms in this way, and observe the nice cancellations.

For example, here is the product of the first seven terms: $$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\frac{8}{7}\frac{7}{8}\frac{9}{8}=\frac{9}{2\cdot 8}.$$

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Let $g(k)=\frac{k-1}{k}$. Then this product is:

$$\prod_{k=2}^n \frac{g(k)}{g(k+1)}$$

which is a telescoping product, and thus equal to $\frac{g(2)}{g(n+1)}$

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$$\left(1-\frac1{(k-1)^2}\right)\left(1-\frac1{(k)^2}\right)\left(1-\frac1{(k+1)^2}\right)$$

$$=\frac{k(k-2)(k-1)(k+1)k(k+2)}{(k-1)^2k^2(k+1)^2}=\frac{(k-2)(k+2)}{(k-1)(k+1)}$$

Observe that $\left(1-\frac1{(k)^2}\right)$ 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 $=\frac{n+1}n$

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