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Unformatted text preview: (2 k3)!! I 2 k2 ² =π 4 n X k =1 1 k 2 And so we have (2 n )!! (2 n1)!! I 2 n = π 3 24π 4 n X k =1 1 k 2 = π 4 ± π 2 6n X k =1 1 k 2 ² 5. State why it is suﬃcient to show that lim n →∞ (2 n )!! (2 n1)!! I 2 n = 0 in order to conclude that ∑ ∞ n =1 1 /n 2 = π 2 / 6. 6. Verify that I 2 n = Z π/ 2 t 2 cos 2 n ( t ) dt ≤ ± π 2 ² 2 Z π/ 2 sin 2 ( t ) cos 2 n ( t ) dt ± π 2 ² 2 Z π/ 2 sin 2 ( t ) cos 2 n ( t ) dt = ± π 2 ² 2 ³Z π/ 2 cos 2 n ( t ) dtZ π/ 2 cos 2 n +2 ( t ) dt ´ 7. Show that ± π 2 ² 2 ³Z π/ 2 cos 2 n ( t ) dtZ π/ 2 cos 2 n +2 ( t ) dt ´ = π 3 8 ³ (2 n1)!! (2 n )!!(2 n + 1)!! (2 n + 2)!! ´ = π 3 8 (2 n1)!! (2 n + 2)!! 8. Deduce that < (2 n )!! (2 n1)!! I 2 n ≤ π 3 8 1 2 n + 2 9. Why are we done?...
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 Spring '08
 Bonner
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