If a₁ = 1 and for ∀ n ≥ 1

a_n+1 = (1/2)a_n + (n² − 2n − 1) / (n²(n+1)²)

Then | Σ_n=1^∞ (a_n − 2/n²) | is equal to

The correct answer is 2.

We are given the recurrence relation for $a_n$ for $n���1$:
$a_{n+1}=\frac{1}{2}{a}_n+\frac{n^2-2n-1}{n^2{(n+1)}^2}$
with the initial term $a_1=1$.

First, let's simplify the fractional term on the right-hand side. We can rewrite the numerator $n^2-2n-1$ as follows:
$n^2-2n-1=2n^2-(n^2+2n+1)=2n^2-{(n+1)}^2$

Using this representation, we can decompose the fraction:
$\frac{n^2-2n-1}{n^2{(n+1)}^2}=\frac{2n^2-{(n+1)}^2}{n^2{(n+1)}^2}=\frac{2}{{(n+1)}^2}-\frac{1}{n^2}$

Now, substitute this back into the recurrence relation:
$a_{n+1}=\frac{1}{2}{a}_n+\frac{2}{{(n+1)}^2}-\frac{1}{n^2}$

To eliminate the coefficient $\frac{1}{2}$ before $a_n$, we multiply the entire equation by $2^{n+1}$:
$2^{n+1}{a}_{n+1}=2^n{a}_n+\frac{2^{n+2}}{{(n+1)}^2}-\frac{2^{n+1}}{n^2}$

We can rearrange this equation to group terms of index $n+1$ and index $n$:
$2^{n+1}{a}_{n+1}-\frac{2^{n+2}}{{(n+1)}^2}=2^n{a}_n-\frac{2^{n+1}}{n^2}$

Let us define a sequence $b_n=2^n{a}_n-\frac{2^{n+1}}{n^2}$. The relation simplifies to:
$b_{n+1}=b_n$
This means $b_n$ is a constant sequence for all $n\ge 1$.

We can find the value of this constant using the initial term $a_1=1$:
$b_1=2^1{a}_1-\frac{2^2}{1^2}=2\left(1\right)-4=-2$

Therefore, for all $n\ge 1$:
$2^n{a}_n-\frac{2^{n+1}}{n^2}=-2$

Divide this equation by $2^n$ to solve for $a_n-\frac{2}{n^2}$:
$a_n-\frac{2}{n^2}=-\frac{2}{2^n}=-\frac{1}{2^{n-1}}$

Next, we sum this expression from $n=1$ to $\infty$:
$\sum _{n=1}^{\infty }(a_n-\frac{2}{n^2})=\sum _{n=1}^{\infty }(-\frac{1}{2^{n-1}})$

This is an infinite geometric series with first term $a=-1$ and common ratio $r=\frac{1}{2}$:
$\sum _{n=1}^{\infty }(-\frac{1}{2^{n-1}})=-(1+\frac{1}{2}+\frac{1}{4}+\cdots )=\frac{-1}{1-\frac{1}{2}}=-2$

Taking the absolute value as requested:
$\left|\sum _{n=1}^{\infty }(a_n-\frac{2}{n^2})\right|=|-2|=2$