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Alternating-parity sum of reciprocal squares

High school · ●●●○○ 50 pts

Compute the value of 1+132+152+172+⋯+19972+19992−110022−110042−110062−⋯−119982−1200021+122+132+142+⋯+19992+110002\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{997^2}+\frac{1}{999^2}-\frac{1}{1002^2}-\frac{1}{1004^2}-\frac{1}{1006^2}-\cdots-\frac{1}{1998^2}-\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{999^2}+\frac{1}{1000^2}} without using a calculator. Give your answer as a decimal.

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