The problem you shared asks to simplify the expression $P = \prod_{k=2}^{n} \left( 1 - \frac{1}{k^2} \right)$, where the product is taken from $k = 2$ to $n$.

Let's begin by simplifying the individual term:

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

Thus, the product becomes:

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

We can break this into two products:

$P = \prod_{k=2}^{n} \frac{(k - 1)}{k} \times \prod_{k=2}^{n} \frac{(k + 1)}{k}$

These two products can now be written as telescoping products. Let's handle each product separately:

1. For $\prod_{k=2}^{n} \frac{k - 1}{k}$, the terms telescope to:

$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{n-1}{n} = \frac{1}{n}$

2. For $\prod_{k=2}^{n} \frac{k + 1}{k}$, the terms telescope to:

$\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{n+1}{n} = \frac{n+1}{2}$

Now, multiplying both simplified products together:

$P = \frac{1}{n} \times \frac{n+1}{2} = \frac{n+1}{2n}$

Thus, the simplified expression for $P$ is:

$P = \frac{n+1}{2n}$

Would you like further clarification or details on any step?

Here are 5 related questions to expand the topic:

1. How would the product change if the starting value of $k$ were different?
2. What happens to $P$ as $n \to \infty$?
3. Can you generalize this product for different powers of $k$?
4. How does the result change if instead of $k^2$, the denominator had $k^3$?
5. Can you apply similar methods to simplify sums instead of products?

Tip: Telescoping series and products are a powerful tool in simplifications, as many intermediate terms cancel out, leaving just a few terms from the beginning and end of the expression.