Let a₁=1 and for n ≥ 1, aₙ₊₁ = 1/2aₙ + (n²-2n-1)/n²(n+1)². Then |… | JEE Main 2026 Mathematics PYQ

NVAMediumJEE Main 2026 · 21 January, Shift 1Sum of Series

Let $a_1=1$ and for $n \ge 1$ , $a_{n+1} = \frac{1}{2}a_n + \frac{n^2-2n-1}{n^2(n+1)^2}$ . Then $\left|\sum_{n=1}^\infty \left(a_n - \frac{2}{n^2}\right)\right|$ is equal to:

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