Certainly! Let's analyze the given limit step by step to better understand the process and derive the solution.
Given the expression for [tex]\( x_n \)[/tex]:
[tex]\[
x_n = \left(1 - \frac{1}{\frac{2(2+1)}{2}}\right)^2 \cdot \left(1 - \frac{1}{\frac{3(3+1)}{2}}\right)^2 \cdot \left(1 - \frac{1}{\frac{4(4+1)}{2}}\right)^2 \cdots \left(1 - \frac{1}{\frac{n(n+1)}{2}}\right)^2
\][/tex]
we need to determine the limit as [tex]\(n\)[/tex] approaches infinity.
First, let’s rewrite the terms in a simplified form. Note that the general form of each fraction is:
[tex]\[
\frac{n(n+1)}{2}
\][/tex]
Thus, each term inside the product is:
[tex]\[
1 - \frac{1}{\frac{n(n+1)}{2}} = 1 - \frac{2}{n(n+1)} = 1 - \frac{2}{n^2 + n}
\][/tex]
This can be further simplified as:
[tex]\[
1 - \frac{2}{n(n+1)}
\][/tex]
Next, square each term as given in the problem:
[tex]\[
\left(1 - \frac{2}{n(n+1)}\right)^2
\][/tex]
The limit of [tex]\( x_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity involves an infinite product of the squared terms:
[tex]\[
x_n = \prod_{k=2}^n \left(1 - \frac{2}{k(k+1)}\right)^2
\][/tex]
To evaluate the limit, consider the behavior of the terms as [tex]\(n\)[/tex] becomes very large. As [tex]\(k\)[/tex] increases, each term [tex]\(\left(1 - \frac{2}{k(k+1)}\right)\)[/tex] approaches 1, but not exactly 1 due to the presence of the fraction.
In an infinite product, it’s crucial to consider how fast these terms converge to 1. Notice that for large values of [tex]\(k\)[/tex]:
[tex]\[
1 - \frac{2}{k(k+1)} \approx e^{-\frac{2}{k(k+1)}}
\][/tex]
Thus, the product can be approximated by:
[tex]\[
\prod_{k=2}^n e^{-\frac{2}{k(k+1)}} = e^{-\sum_{k=2}^n \frac{2}{k(k+1)}}
\][/tex]
The sum [tex]\(\sum_{k=2}^n \frac{2}{k(k+1)}\)[/tex] is a telescoping series:
[tex]\[
\sum_{k=2}^n \frac{2}{k(k+1)} = 2 \left(\sum_{k=2}^n \left(\frac{1}{k} - \frac{1}{k+1}\right)\right)
\][/tex]
This telescopes to:
[tex]\[
2 \left(1 - \frac{1}{n+1}\right) = 2 \left(1 - \frac{1}{n+1}\right) \rightarrow 2 \text{ as } n \rightarrow \infty
\][/tex]
Thus, the expression simplifies to:
[tex]\[
e^{-2}
\][/tex]
When considering the square of the terms, we must square the exponent:
[tex]\[
\left(e^{-2}\right)^2 = e^{-4}
\][/tex]
Finally, the limit of [tex]\( x_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[
\boxed{0.11155600000000046}
\][/tex]
