We are given the recurrence relation $a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$ with $a_1 = 1$. We want to find the value of $\sum_{n=1}^{\infty} (a_n - \frac{2}{n^2})$.

First, let's rewrite the recurrence relation as:

$a_{n+1} = \frac{1}{2}a_n + \frac{(n+1)^2 - 4n - 2}{n^2(n+1)^2} = \frac{1}{2}a_n + \frac{1}{n^2} - \frac{4n+2}{n^2(n+1)^2}$

Let $b_n = a_n - \frac{2}{n^2}$. Then $a_n = b_n + \frac{2}{n^2}$. Substituting this into the recurrence relation:

$b_{n+1} + \frac{2}{(n+1)^2} = \frac{1}{2}(b_n + \frac{2}{n^2}) + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$

$b_{n+1} = \frac{1}{2}b_n + \frac{1}{n^2} - \frac{2}{(n+1)^2} + \frac{n^2 - 2n - 1}{n^2(n+1)^2} - \frac{2}{(n+1)^2} = \frac{1}{2}b_n + \frac{1}{n^2} - \frac{1}{n(n+1)} - \frac{1}{(n+1)^2} - \frac{2}{(n+1)^2}$

$b_{n+1} = \frac{1}{2}b_n + \frac{1}{n^2} - \frac{2}{(n+1)^2} + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$

Let's try to find a telescoping sum. Consider $a_{n+1} - \frac{1}{n^2} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2(n+1)^2} - \frac{1}{n^2} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1 - (n+1)^2}{n^2(n+1)^2} = \frac{1}{2}a_n + \frac{-4n-2}{n^2(n+1)^2} = \frac{1}{2}a_n - \frac{2(2n+1)}{n^2(n+1)^2}$

Consider $c_n = a_n - \frac{2}{n^2}$. Then $a_n = c_n + \frac{2}{n^2}$. Substituting into the recurrence:

$c_{n+1} + \frac{2}{(n+1)^2} = \frac{1}{2}(c_n + \frac{2}{n^2}) + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$

$c_{n+1} = \frac{1}{2}c_n + \frac{1}{n^2} - \frac{2}{(n+1)^2} + \frac{n^2 - 2n - 1}{n^2(n+1)^2} = \frac{1}{2}c_n + \frac{(n+1)^2 - 2n^2 + n^2 - 2n - 1}{n^2(n+1)^2} = \frac{1}{2}c_n + \frac{0}{n^2(n+1)^2} = \frac{1}{2}c_n$

So $c_{n+1} = \frac{1}{2}c_n$. This means $c_n$ is a geometric progression with ratio $\frac{1}{2}$.

We have $c_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$. Thus $c_n = -(\frac{1}{2})^{n-1}$.

We want to find $\sum_{n=1}^{\infty} c_n = \sum_{n=1}^{\infty} (a_n - \frac{2}{n^2}) = \sum_{n=1}^{\infty} -(\frac{1}{2})^{n-1} = -\sum_{n=0}^{\infty} (\frac{1}{2})^n = -\frac{1}{1 - \frac{1}{2}} = -2$.