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

1. Problem Description

The problem asks 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 of the series:

$\sum_{k=2}^{\infty} \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right) = \left( \frac{3}{(2-1)^2} - \frac{3}{2^2} \right) + \left( \frac{3}{(3-1)^2} - \frac{3}{3^2} \right) + \left( \frac{3}{(4-1)^2} - \frac{3}{4^2} \right) + \dots$

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

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

Let $S_n$ be the partial sum of the first $n$ terms.

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

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

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

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

Now, we take the limit as $n$ approaches infinity:

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 3 - \frac{3}{(n+1)^2} \right)$

$= 3 - \lim_{n \to \infty} \frac{3}{(n+1)^2}$

Since $\lim_{n \to \infty} (n+1)^2 = \infty$, we have

$= 3 - 0 = 3$

3. Final Answer

$3$