14-15 - SOLUTIONS FOR HOMEWORK 7 14.1. (a) an = n4 /2n . We...

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SOLUTIONS FOR HOMEWORK 7 14.1. (a) a n = n 4 / 2 n . We know that lim n 1 /n = 1 (Example 9.7). There- fore, lim | a n | 1 /n = 1 / 2 < 1. By Root Test, the series converges. (b) a n = 2 n /n !. lim( n !) 1 /n = + (proved in class), hence lim | a n | 1 /n = 0. By Root Test, the series converges. Alternatively , note that | a n +1 | / | a n | = 2 / ( n + 1), hence lim | a n +1 | / | a n | = 0. By Ratio Test, our series converges. (e) 0 a n = cos 2 n/n 2 b n = 1 /n 2 . b n converges (Integral Test), hence, by Comparison Test, a n converges. 14.4. (a) a n = 1 / ( n + ( 1) n ) 2 . The series n a n converges. For n 2, 0 a n b n = 1 / ( n 1) 2 , hence | a n | ≤ b n . However, n b n converges (Example 2 from p. 69), hence, by Comparison Test, n a n also converges. (b) a n = n + 1 n . The series n a n diverges to + . To see this, consider the sequence of partial sums: s n = n k =1 a k = n k =1 ( k + 1 k ) = n + 1 1 (this is a “telescopic sum”). Thus, lim s n = + . Alternatively , one could observe that a n = ( n + 1 n ) · n + 1 + n n + 1 + n = 1 n + 1 + n > 1 2 n + 1 = b n . However, n b n diverges, by Integral Test. (c) [ This is a bonus problem – very little partial credit is given ] a n = n ! /n n . The series n a n converges. For instance, one can apply Root Test. Note ±rst (it was probably shown in the discussion) that, for any n N , (1) n ! ( n/ 2) n/ 2 1 · n n/ 2+1 = 2 n n 2 n . Indeed, consider ±rst the case of n even, that is, n = 2 m . Then n ! = ( 1 · 2 · . . . · m ) · ( ( m + 1) · . . . · n ) . Then 1
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14-15 - SOLUTIONS FOR HOMEWORK 7 14.1. (a) an = n4 /2n . We...

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