If $S = \lim_{n o \infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$, then S is equal to:

Question

If $S = \lim_{n 	o \infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$, then S is equal to:

Prepp Team · Accepted Answer

Evaluating the Infinite Product Limit S

The question asks us to find the value of the limit S, where S is defined as the infinite product:

$S = \lim_{n \to \infty} \left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{3^2}\right) \left(1-\frac{1}{4^2}\right) \dots \left(1-\frac{1}{n^2}\right)$

Let's denote the partial product up to n terms as $P_n$:

$P_n = \prod_{k=2}^{n} \left(1-\frac{1}{k^2}\right)$

We need to evaluate $S = \lim_{n \to \infty} P_n$. Let's simplify the general term inside the product first.

Step 1: Simplify the General Term

The general term is $\left(1-\frac{1}{k^2}\right)$. We can rewrite this using algebraic manipulation:

$1-\frac{1}{k^2} = \frac{k^2}{k^2} - \frac{1}{k^2} = \frac{k^2 - 1}{k^2}$

Further factorizing the numerator using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$):

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

Step 2: Express the Partial Product $P_n$

Now, substitute the simplified term back into the expression for $P_n$:

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

Let's write out the first few terms and the last term to see the pattern:

$P_n = \frac{(2-1)(2+1)}{2 \cdot 2} \times \frac{(3-1)(3+1)}{3 \cdot 3} \times \frac{(4-1)(4+1)}{4 \cdot 4} \times \dots \times \frac{(n-1)(n+1)}{n \cdot n}$ $P_n = \frac{1 \cdot 3}{2 \cdot 2} \times \frac{2 \cdot 4}{3 \cdot 3} \times \frac{3 \cdot 5}{4 \cdot 4} \times \dots \times \frac{(n-1)(n+1)}{n \cdot n}$

Step 3: Simplify the Product using Cancellation

This type of product is called a telescoping product. We can rearrange the terms to observe the cancellations:

$P_n = \left( \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{n-1}{n} \right) \times \left( \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{n+1}{n} \right)$

Let's evaluate the two parts separately:

The first part is: $ \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{n-1}{n} $ Notice that the numerator of each fraction cancels the denominator of the previous fraction. After cancellation, we are left with: $ \frac{1}{n} $
The second part is: $ \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{n+1}{n} $ Here, the numerator of each fraction cancels the denominator of the *next* fraction (or equivalently, the denominator cancels the numerator of the previous one). Let's write it explicitly: $ \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \ldots \cdot \frac{n}{n-1} \cdot \frac{n+1}{n} $ After cancellation (3s, 4s, ..., ns), we are left with: $ \frac{n+1}{2} $

Now, multiply the results of the two parts:

$P_n = \left( \frac{1}{n} \right) \times \left( \frac{n+1}{2} \right) = \frac{n+1}{2n}$

Step 4: Calculate the Limit

Finally, we need to find the limit of $P_n$ as $n$ approaches infinity:

$S = \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{n+1}{2n}$

To evaluate this limit, we can divide both the numerator and the denominator by the highest power of $n$, which is $n$:

$S = \lim_{n \to \infty} \frac{\frac{n}{n} + \frac{1}{n}}{\frac{2n}{n}} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2}$

As $n \to \infty$, the term $\frac{1}{n}$ approaches 0:

$S = \frac{1 + 0}{2} = \frac{1}{2}$

Conclusion

The value of the infinite product limit S is $\frac{1}{2}$.

Prepp Team · Answer

0

Prepp Team · Answer

$\frac{1}{4}$

Prepp Team · Answer

1