# Limit Involving a Product of a Sequence: $\lim _{x\to\infty }\left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\dots\left(1-\frac1{x^2}\right)$ [duplicate]

I am having trouble figuring out how to solve this limit.

$$\lim _{x\to \infty }\left(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{x^2}\right)\right)$$

I understand that as '$x$' increases, the overall product becomes an even smaller number between $0$ and $1$ only because I tried plugging in numbers in hope to learn about the nature of the expression. But I can't seem to come to a pattern that will allow me to efficiently simplify it and solve the limit.

Help, anyone?

## marked as duplicate by Martin Sleziak, Jonas Meyer, Community♦Jul 28 '16 at 20:52

• I think the first factor is $1-\frac1{2^2}$ not $1-\frac1{1^2}$, the latter being equal to $0$. – Olivier Oloa Jul 28 '16 at 20:22
• Should the first term 1 - 1/1 not be there? Since the whole expression is rendered 0 because of that. – zarathustra Jul 28 '16 at 20:23
• yes you are right, I'll fix it right away, sorry. – intersomnium Jul 28 '16 at 20:23

Overkill: $$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$ gives: $$\prod_{n\geq 2}\left(1-\frac{1}{n^2}\right) = \lim_{z \to 1}\frac{\sin(\pi z)}{\pi z(1-z^2)} \stackrel{DH}{=}\lim_{z\to 1}\frac{\cos(\pi z)}{1-3z^2}=\color{red}{\frac{1}{2}}.\tag{2}$$ That stops being an overkill if the infinite product you want to compute is $\prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right)=\frac{2}{\pi}$ (aka Wallis' product), for instance.

• What does the "DH" above the equals sign stand for? – Jonas Meyer Jul 28 '16 at 20:44
• @JonasMeyer: De l'Hopital rule. – Jack D'Aurizio Jul 28 '16 at 20:45

Hint. One may observe that $$\prod_{n=2}^N\left(1-\frac1{n^2}\right)=\prod_{n=2}^N\frac{n^2-1}{n^2}=\prod_{n=2}^N\frac{n+1}{n}\cdot \prod_{n=2}^N\frac{n-1}{n}$$ then factors telescope.

• yeah,i also suspected it – haqnatural Jul 28 '16 at 20:23

The logarithm of the limit is the limit of the logarithm, now:

$$\lim_{n \to \infty} \sum_{k=2}^{n} \log \left( 1 - \frac1{k^2}\right) = \lim_{n \to \infty} \left[ \sum_{k=2}^{n} (\log (k+1) - \log k) - \sum_{k=2}^n (\log k - \log (k-1)) \right] = \lim_{n \to \infty} \left[ \log\left( \frac{n+1}n \right) - \log 2 \right] = - \log 2$$

So the answer is $1/2$.