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n = 1 1 2 ( n + 1 )

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n = 1 1 2 n 1

Diverges by comparison with harmonic series, since 2 n 1 n .

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n = 2 1 ( n ln n ) 2

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n = 1 n ! ( n + 2 ) !

a n = 1 / ( n + 1 ) ( n + 2 ) < 1 / n 2 . Converges by comparison with p -series, p = 2 .

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n = 1 sin ( 1 / n ) n

sin ( 1 / n ) 1 / n , so converges by comparison with p -series, p = 2 .

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n = 1 sin 2 n n 2

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n = 1 sin ( 1 / n ) n

sin ( 1 / n ) 1 , so converges by comparison with p -series, p = 3 / 2 .

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n = 1 n 1.2 1 n 2.3 + 1

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n = 1 n + 1 n n

Since n + 1 n = 1 / ( n + 1 + n ) 2 / n , series converges by comparison with p -series for p = 1.5.

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n = 1 n 4 n 4 + n 2 3

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Use the limit comparison test to determine whether each of the following series converges or diverges.

n = 1 ( ln n n ) 2

Converges by limit comparison with p -series for p > 1 .

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n = 1 ( ln n n 0.6 ) 2

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n = 1 ln ( 1 + 1 n ) n

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

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n = 1 ln ( 1 + 1 n 2 )

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n = 1 1 4 n 3 n

Converges by limit comparison with 4 n .

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n = 1 1 n 2 n sin n

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n = 1 1 e ( 1.1 ) n 3 n

Converges by limit comparison with 1 / e 1.1 n .

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n = 1 1 e ( 1.01 ) n 3 n

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n = 1 1 n 1 + 1 / n

Diverges by limit comparison with harmonic series.

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n = 1 1 2 1 + 1 / n n 1 + 1 / n

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n = 1 ( 1 n sin ( 1 n ) )

Converges by limit comparison with p -series, p = 3 .

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n = 1 ( 1 cos ( 1 n ) )

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n = 1 1 n ( tan −1 n π 2 )

Converges by limit comparison with p -series, p = 3 .

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n = 1 ( 1 1 n ) n . n ( Hint: ( 1 1 n ) n 1 / e . )

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n = 1 ( 1 e −1 / n ) ( Hint: 1 / e ( 1 1 / n ) n , so 1 e −1 / n 1 / n . )

Diverges by limit comparison with 1 / n .

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Does n = 2 1 ( ln n ) p converge if p is large enough? If so, for which p ?

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Does n = 1 ( ( ln n ) n ) p converge if p is large enough? If so, for which p ?

Converges for p > 1 by comparison with a p series for slightly smaller p .

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For which p does the series n = 1 2 p n / 3 n converge?

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For which p > 0 does the series n = 1 n p 2 n converge?

Converges for all p > 0 .

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For which r > 0 does the series n = 1 r n 2 2 n converge?

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For which r > 0 does the series n = 1 2 n r n 2 converge?

Converges for all r > 1 . If r > 1 then r n > 4 , say, once n > ln ( 2 ) / ln ( r ) and then the series converges by limit comparison with a geometric series with ratio 1 / 2 .

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Find all values of p and q such that n = 1 n p ( n ! ) q converges.

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Does n = 1 sin 2 ( n r / 2 ) n converge or diverge? Explain.

The numerator is equal to 1 when n is odd and 0 when n is even, so the series can be rewritten n = 1 1 2 n + 1 , which diverges by limit comparison with the harmonic series.

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Explain why, for each n , at least one of { | sin n | , | sin ( n + 1 ) | ,... , | sin n + 6 | } is larger than 1 / 2 . Use this relation to test convergence of n = 1 | sin n | n .

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Suppose that a n 0 and b n 0 and that n = 1 a 2 n and n = 1 b 2 n converge. Prove that n = 1 a n b n converges and n = 1 a n b n 1 2 ( n = 1 a n 2 + n = 1 b n 2 ) .

( a b ) 2 = a 2 2 a b + b 2 or a 2 + b 2 2 a b , so convergence follows from comparison of 2 a n b n with a 2 n + b 2 n . Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.

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Does n = 1 2 ln ln n converge? ( Hint: Write 2 ln ln n as a power of ln n . )

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Does n = 1 ( ln n ) ln n converge? ( Hint: Use t = e ln ( t ) to compare to a p series . )

( ln n ) ln n = e ln ( n ) ln ln ( n ) . If n is sufficiently large, then ln ln n > 2 , so ( ln n ) ln n < 1 / n 2 , and the series converges by comparison to a p series .

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Does n = 2 ( ln n ) ln ln n converge? ( Hint: Compare a n to 1 / n . )

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Show that if a n 0 and n = 1 a n converges, then n = 1 a 2 n converges. If n = 1 a 2 n converges, does n = 1 a n necessarily converge?

a n 0 , so a 2 n | a n | for large n . Convergence follows from limit comparison. 1 / n 2 converges, but 1 / n does not, so the fact that n = 1 a 2 n converges does not imply that n = 1 a n converges.

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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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