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Question Number 203846 by York12 last updated on 30/Jan/24

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{i=2}{\prod ^n}\left(\frac{i^2-1}{i^2}\right)$$

Answered by MM42 last updated on 30/Jan/24

$$p_n=\underset{i=2}{\prod ^n}\frac{i^2-1}{i^2}=\underset{i=2}{\prod ^n}\frac{(i-1)(i+1)}{i^2}$$$$\Rightarrow p_n=\frac{\left(1\times 3\right)\left(2\times 4\right)\left(3\times 5\right)...(n-1)(n+1)}{2^2\times 3^2\times 4^2\times ...\times n^2}$$$$=\frac{n+1}{2n}\Rightarrow {lim}_{n\to \mathrm{\infty }}{p}_n=\frac{1}{2}✓$$$$$$

Commented by York12 last updated on 30/Jan/24

$$\mathrm{thanks}$$

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