\[ \large \sum_{k=1}^n \frac 1 {k^2 x - 1} = \frac 1 2 \]

Consider the equation above, for every natural number \(n\), this equation admits a unique solution \(x_n > 1 \) (One can prove this).

Suppose the sequence \(x_n \) converges to \(l\), find \(l^{2}\).

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