We are asked to evaluate the infinite sum $\sum_{k=2}^\infty \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right)$. This looks like a telescoping series.

1. Problem Description

We are asked to evaluate the infinite sum $\sum_{k=2}^\infty \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right)$. This looks like a telescoping series.

2. Solution Steps

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

When $k=2$, we have $\frac{3}{(2-1)^2} - \frac{3}{2^2} = \frac{3}{1^2} - \frac{3}{2^2} = 3 - \frac{3}{4}$.

When $k=3$, we have $\frac{3}{(3-1)^2} - \frac{3}{3^2} = \frac{3}{2^2} - \frac{3}{3^2} = \frac{3}{4} - \frac{3}{9}$.

When $k=4$, we have $\frac{3}{(4-1)^2} - \frac{3}{4^2} = \frac{3}{3^2} - \frac{3}{4^2} = \frac{3}{9} - \frac{3}{16}$.

When $k=5$, we have $\frac{3}{(5-1)^2} - \frac{3}{5^2} = \frac{3}{4^2} - \frac{3}{5^2} = \frac{3}{16} - \frac{3}{25}$.

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

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

We see that this is a telescoping sum, where consecutive terms cancel each other out.

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

Now we need to find the limit as $n \to \infty$.

$\sum_{k=2}^\infty \left( \frac{3}{(k-1)^2} - \frac{3}{k^2} \right) = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 3 - \frac{3}{(n+1)^2} \right) = 3 - 0 = 3$.

3. Final Answer

$3$