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I a 2n -3.

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On pose PnSn On dmontre que Pn1 Sn1 Donc uou1un 2- 2n32n uou1un un1 2- 2n32n un1 Sn1 Mais aprs je coince un peu daide nest pas de. 1 we will prove that the statement must be true for n k 1. 4prouver alors qie la suite u n converge vers ln2.
Here is all you have to to learn about 2 3 1 n 1 2n X x 1 x 3 x 6 x x 2 x 3 x 5 x x 3 x 5 x 6 n 1 n 1 n 1 Si n impair. Car alors produit de deux pairs successifs dont lun est divisible par 4. Since n4 1 n4 we have 1 n41 1 n4 so a n n 2 n4 1 n n4 1 n2 therefore 0 n 1 n2 Since the p-series P 1 n1 1 2 converges the comparison test tells us that the series P 1 n1 n2 n41 converges also. 0 1 2 M m 13.
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