JEE Main 202621 Jan, Morning shiftMathematicsMediumSequence and Series179× since 2002Let a1=1a_1=1a1=1 and for n≥1n\ge1n≥1, an+1=12an+n2−2n−1n2(n+1)2a_{n+1}=\frac12a_n+\frac{n^2-2n-1}{n^2(n+1)^2}an+1=21an+n2(n+1)2n2−2n−1. Then ∣∑n=1∞(an−2n2)∣\left|\sum_{n=1}^{\infty}\left(a_n-\frac{2}{n^2}\right)\right|∑n=1∞(an−n22) is equal to ______.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
