We need to evaluate the infinite sum $\sum_{k=2}^{\infty} \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right)$.

1. Problem Description

We need to evaluate the infinite sum $\sum_{k=2}^{\infty} \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right)$.

2. Solution Steps

This is a telescoping series. Let's write out the first few terms:

$k=2: \frac{3}{(2-1)^2} - \frac{3}{2^2} = \frac{3}{1} - \frac{3}{4}$

$k=3: \frac{3}{(3-1)^2} - \frac{3}{3^2} = \frac{3}{4} - \frac{3}{9}$

$k=4: \frac{3}{(4-1)^2} - \frac{3}{4^2} = \frac{3}{9} - \frac{3}{16}$

$k=5: \frac{3}{(5-1)^2} - \frac{3}{5^2} = \frac{3}{16} - \frac{3}{25}$

We can see that the terms are cancelling each other out.

Let $S_n = \sum_{k=2}^{n} \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right)$. Then

$S_n = \left( \frac{3}{1^2} - \frac{3}{2^2} \right) + \left( \frac{3}{2^2} - \frac{3}{3^2} \right) + \left( \frac{3}{3^2} - \frac{3}{4^2} \right) + ... + \left( \frac{3}{(n-1)^2} - \frac{3}{n^2} \right)$

$S_n = \frac{3}{1^2} - \frac{3}{n^2} = 3 - \frac{3}{n^2}$

As $n$ approaches infinity, $\frac{3}{n^2}$ approaches

0. Therefore, $\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 3 - \frac{3}{n^2} \right) = 3 - 0 = 3$.

3. Final Answer

$3$