x(1-1%2F3%5E2)x(1-1%2F4%5E2)-to-infinity.jpeg "(1-1/2^2)x(1-1/3^2)x(1-1/4^2) To Infinity")
Infinite Product: (1-1/2^2)x(1-1/3^2)x(1-1/4^2) to Infinity
In mathematics, infinite products are used to represent a product of an infinite number of terms. One such infinite product is:
$(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdots$
where the expression continues indefinitely. In this article, we will explore this infinite product and its properties.
Simplifying the Expression
To simplify the expression, we can start by noting that:
$1-\frac{1}{n^2} = \frac{n^2-1}{n^2} = \frac{(n-1)(n+1)}{n^2}$
Using this formula, we can rewrite the infinite product as:
$\frac{(1-1)(1+1)}{2^2} \cdot \frac{(2-1)(2+1)}{3^2} \cdot \frac{(3-1)(3+1)}{4^2} \cdots$
Evaluating the Product
To evaluate the infinite product, we can start by canceling out the common terms:
$\frac{1 \cdot 2}{2^2} \cdot \frac{2 \cdot 3}{3^2} \cdot \frac{3 \cdot 4}{4^2} \cdots$
As we continue the product, we can see that all terms cancel out, except for the numerator of the first term and the denominator of the last term. Therefore, the infinite product simplifies to:
$\frac{1}{n}$
where $n$ is an arbitrary large integer.
Conclusion
The infinite product $(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdots$ simplifies to $\frac{1}{n}$, where $n$ is an arbitrary large integer. This result has important implications in number theory and algebra.
