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Q54Section C (Section C)54 of 60

Telescoping series in reciprocals of squares

Let a1=1a_1=1 and an=2−1na_n=2-\frac1n for n≥2n\ge2. Then ∑n=1∞(1an2−1an+12)\sum_{n=1}^\infty\left(\frac1{a_n^2}-\frac1{a_{n+1}^2}\right) converges to ________________________

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