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n = 1 ( n ! ) 3 ( 3 n ! )

a n + 1 a n 1 / 27 < 1 . Converges.

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n = 1 2 3 n ( n ! ) 3 ( 3 n ! )

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n = 1 ( 2 n ) ! n 2 n

a n + 1 a n 4 / e 2 < 1 . Converges.

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n = 1 ( 2 n ) ! ( 2 n ) n

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n = 1 n ! ( n / e ) n

a n + 1 a n 1 . Ratio test is inconclusive.

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n = 1 ( 2 n ) ! ( n / e ) 2 n

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n = 1 ( 2 n n ! ) 2 ( 2 n ) 2 n

a n a n + 1 1 / e 2 . Converges.

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Use the root test to determine whether n = 1 a n converges, where a n is as follows.

a k = ( k 1 2 k + 3 ) k

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a k = ( 2 k 2 1 k 2 + 3 ) k

( a k ) 1 / k 2 > 1 . Diverges.

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a n = n / 2 n

( a n ) 1 / n 1 / 2 < 1 . Converges.

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a k = k e e k

( a k ) 1 / k 1 / e < 1 . Converges.

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a n = ( 1 e + 1 n ) n

a n 1 / n = 1 e + 1 n 1 e < 1 . Converges.

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a n = ( ln ( 1 + ln n ) ) n ( ln n ) n

a n 1 / n = ( ln ( 1 + ln n ) ) ( ln n ) 0 by L’Hôpital’s rule. Converges.

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In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series k = 1 a k with given terms a k converges, or state if the test is inconclusive.

a k = k ! 1 · 3 · 5 ( 2 k 1 )

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a k = 2 · 4 · 6 2 k ( 2 k ) !

a k + 1 a k = 1 2 k + 1 0 . Converges by ratio test.

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a k = 1 · 4 · 7 ( 3 k 2 ) 3 k k !

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a n = ( 1 1 n ) n 2

( a n ) 1 / n 1 / e . Converges by root test.

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a k = ( 1 k + 1 + 1 k + 2 + + 1 2 k ) k ( Hint: Compare a k 1 / k to k 2 k d t t . )

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a k = ( 1 k + 1 + 1 k + 2 + + 1 3 k ) k

a k 1 / k ln ( 3 ) > 1 . Diverges by root test.

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a n = ( n 1 / n 1 ) n

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Use the ratio test to determine whether n = 1 a n converges, or state if the ratio test is inconclusive.

n = 1 3 n 2 2 n 3

a n + 1 a n = 3 2 n + 1 2 3 n 2 + 3 n + 1 0 . Converge.

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n = 1 2 n 2 n n n !

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Use the root and limit comparison tests to determine whether n = 1 a n converges.

a n = 1 / x n n where x n + 1 = 1 2 x n + 1 x n , x 1 = 1 ( Hint: Find limit of { x n } . )

Converges by root test and limit comparison test since x n 2 .

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In the following exercises, use an appropriate test to determine whether the series converges.

n = 1 ( n + 1 ) n 3 + n 2 + n + 1

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n = 1 ( −1 ) n + 1 ( n + 1 ) n 3 + 3 n 2 + 3 n + 1

Converges absolutely by limit comparison with p series, p = 2 .

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n = 1 ( n + 1 ) 2 n 3 + ( 1.1 ) n

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n = 1 ( n 1 ) n ( n + 1 ) n

lim n a n = 1 / e 2 0 . Series diverges.

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a n = ( 1 + 1 n 2 ) n ( Hint: ( 1 + 1 n 2 ) n 2 e . )

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a k = 1 / 2 sin 2 k

Terms do not tend to zero: a k 1 / 2 , since sin 2 x 1 .

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a k = 2 sin ( 1 / k )

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a n = 1 / ( n + 2 n ) where ( n k ) = n ! k ! ( n k ) !

a n = 2 ( n + 1 ) ( n + 2 ) , which converges by comparison with p series for p = 2 .

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a k = 2 k / ( 3 k k )

a k = 2 k 1 · 2 k ( 2 k + 1 ) ( 2 k + 2 ) 3 k ( 2 / 3 ) k converges by comparison with geometric series.

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a k = ( k k + ln k ) k ( Hint: a k = ( 1 + ln k k ) ( k / ln k ) ln k e ln k . )

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a k = ( k k + ln k ) 2 k ( Hint: a k = ( 1 + ln k k ) ( k / ln k ) ln k 2 . )

a k e ln k 2 = 1 / k 2 . Series converges by limit comparison with p series, p = 2 .

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The following series converge by the ratio test. Use summation by parts, k = 1 n a k ( b k + 1 b k ) = [ a n + 1 b n + 1 a 1 b 1 ] k = 1 n b k + 1 ( a k + 1 a k ) , to find the sum of the given series.

k = 1 k 2 k ( Hint: Take a k = k and b k = 2 1 k . )

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k = 1 k c k , where c > 1 ( Hint: Take a k = k and b k = c 1 k / ( c 1 ) . )

If b k = c 1 k / ( c 1 ) and a k = k , then b k + 1 b k = c k and n = 1 k c k = a 1 b 1 + 1 c 1 k = 1 c k = c ( c 1 ) 2 .

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n = 1 ( n + 1 ) 2 2 n

6 + 4 + 1 = 11

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The k th term of each of the following series has a factor x k . Find the range of x for which the ratio test implies that the series converges.

k = 1 x 2 k k 2

| x | 1

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k = 1 x k k !

| x | <

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Does there exist a number p such that n = 1 2 n n p converges?

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Let 0 < r < 1 . For which real numbers p does n = 1 n p r n converge?

All real numbers p by the ratio test.

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Suppose that lim n | a n + 1 a n | = p . For which values of p must n = 1 2 n a n converge?

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Suppose that lim n | a n + 1 a n | = p . For which values of r > 0 is n = 1 r n a n guaranteed to converge?

r < 1 / p

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Suppose that | a n + 1 a n | ( n + 1 ) p for all n = 1 , 2 ,… where p is a fixed real number. For which values of p is n = 1 n ! a n guaranteed to converge?

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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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