To determine whether the given expression converges or diverges as $n$ approaches infinity, we will analyze and simplify it:

Given expression:

$\frac{\frac{2}{n} \left(\frac{1}{n} - 3\right)}{\left(\frac{1}{n} - 1\right)^2 + \left(\frac{2}{n}\right) - 1}$

Step 1: Simplify the numerator and denominator

1. Simplify the numerator: $\frac{2}{n} \left(\frac{1}{n} - 3\right)$ Rewrite the expression: $\frac{2}{n} \times \left(\frac{1}{n} - 3\right) = \frac{2 \times 1}{n^2} - \frac{6}{n} = \frac{2}{n^2} - \frac{6}{n}$

2. Simplify the denominator: $\left(\frac{1}{n} - 1\right)^2 + \frac{2}{n} - 1$

   • The term $\left(\frac{1}{n} - 1\right)^2$ can be expanded as: $\left(\frac{1}{n} - 1\right)^2 = \left(\frac{1}{n} - 1\right)\left(\frac{1}{n} - 1\right) = \frac{1}{n^2} - \frac{2}{n} + 1$
   • The full expression for the denominator is: $\frac{1}{n^2} - \frac{2}{n} + 1 + \frac{2}{n} - 1$ Notice that $-\frac{2}{n}$ and $\frac{2}{n}$ cancel out, leaving: $\frac{1}{n^2} + 0 = \frac{1}{n^2}$

Step 2: Combine the simplified terms

The expression simplifies to:

$\frac{\frac{2}{n^2} - \frac{6}{n}}{\frac{1}{n^2}}$

We can further simplify this by multiplying both the numerator and the denominator by $n^2$:

$\frac{2 - 6n}{1} = 2 - 6n$

Step 3: Analyze the behavior as $n$ approaches infinity

As $n$ increases, the term $-6n$ dominates, causing the expression $2 - 6n$ to become more and more negative without bound.

Conclusion:

Since the expression $2 - 6n$ diverges to negative infinity as $n$ approaches infinity, the given expression diverges.

Would you like more details or have any questions?

Here are 5 related questions to further explore the topic:

1. How does multiplying and simplifying fractions help in determining the limit of an expression?
2. What is the significance of the leading term in determining the behavior of an expression as $n$ approaches infinity?
3. How can the squeeze theorem be applied to determine convergence or divergence?
4. What is the role of polynomial degrees in determining the limits of rational expressions?
5. Can an expression converge if the numerator grows faster than the denominator?

Tip: Always simplify complex expressions by canceling out terms and focusing on the leading terms to determine their behavior as $n$ approaches infinity.