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Suppose that a n > 0 for all n and that n = 1 a n converges. Suppose that b n is an arbitrary sequence of zeros and ones. Does n = 1 a n b n necessarily converge?

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Suppose that a n > 0 for all n and that n = 1 a n diverges. Suppose that b n is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does n = 1 a n b n necessarily diverge?

No. n = 1 1 / n diverges. Let b k = 0 unless k = n 2 for some n . Then k b k / k = 1 / k 2 converges.

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Complete the details of the following argument: If n = 1 1 n converges to a finite sum s , then 1 2 s = 1 2 + 1 4 + 1 6 + and s 1 2 s = 1 + 1 3 + 1 5 + . Why does this lead to a contradiction?

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Show that if a n 0 and n = 1 a 2 n converges, then n = 1 sin 2 ( a n ) converges.

| sin t | | t | , so the result follows from the comparison test.

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Suppose that a n / b n 0 in the comparison test, where a n 0 and b n 0. Prove that if b n converges, then a n converges.

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Let b n be an infinite sequence of zeros and ones. What is the largest possible value of x = n = 1 b n / 2 n ?

By the comparison test, x = n = 1 b n / 2 n n = 1 1 / 2 n = 1.

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Let d n be an infinite sequence of digits, meaning d n takes values in { 0 , 1 ,… , 9 } . What is the largest possible value of x = n = 1 d n / 10 n that converges?

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Explain why, if x > 1 / 2 , then x cannot be written x = n = 2 b n 2 n ( b n = 0 or 1 , b 1 = 0 ) .

If b 1 = 0 , then, by comparison, x n = 2 1 / 2 n = 1 / 2 .

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[T] Evelyn has a perfect balancing scale, an unlimited number of 1 -kg weights, and one each of 1 / 2 -kg , 1 / 4 -kg , 1 / 8 -kg , and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?

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[T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of 1 -kg weights, and nine each of 0.1 -kg, 0.01 -kg , 0.001 -kg, and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?

Yes. Keep adding 1 -kg weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the 1 -kg weights, and add 0.1 -kg weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last 0.1 -kg weight. Start adding 0.01 -kg weights. If it balances, stop. If it tips to the side with the weights, remove the last 0.01 -kg weight that was added. Continue in this way for the 0.001 -kg weights, and so on. After a finite number of steps, one has a finite series of the form A + n = 1 N s n / 10 n where A is the number of full kg weights and d n is the number of 1 / 10 n -kg weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the N th partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most 1 / 10 N .

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The series n = 1 1 2 n is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which n is odd. Let m > 1 be fixed. Show, more generally, that deleting all terms 1 / n where n = m k for some integer k also results in a divergent series.

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In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from n = 1 1 n by removing any term 1 / n if a given digit, say 9 , appears in the decimal expansion of n . Argue that this depleted harmonic series converges by answering the following questions.

  1. How many whole numbers n have d digits?
  2. How many d -digit whole numbers h ( d ) . do not contain 9 as one or more of their digits?
  3. What is the smallest d -digit number m ( d ) ?
  4. Explain why the deleted harmonic series is bounded by d = 1 h ( d ) m ( d ) .
  5. Show that d = 1 h ( d ) m ( d ) converges.

a. 10 d 10 d 1 < 10 d b. h ( d ) < 9 d c. m ( d ) = 10 d 1 + 1 d. Group the terms in the deleted harmonic series together by number of digits. h ( d ) bounds the number of terms, and each term is at most 1 / m ( d ) . d = 1 h ( d ) / m ( d ) d = 1 9 d / ( 10 ) d 1 90. One can actually use comparison to estimate the value to smaller than 80 . The actual value is smaller than 23 .

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Suppose that a sequence of numbers a n > 0 has the property that a 1 = 1 and a n + 1 = 1 n + 1 S n , where S n = a 1 + + a n . Can you determine whether n = 1 a n converges? ( Hint: S n is monotone.)

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Suppose that a sequence of numbers a n > 0 has the property that a 1 = 1 and a n + 1 = 1 ( n + 1 ) 2 S n , where S n = a 1 + + a n . Can you determine whether n = 1 a n converges? ( Hint: S 2 = a 2 + a 1 = a 2 + S 1 = a 2 + 1 = 1 + 1 / 4 = ( 1 + 1 / 4 ) S 1 , S 3 = 1 3 2 S 2 + S 2 = ( 1 + 1 / 9 ) S 2 = ( 1 + 1 / 9 ) ( 1 + 1 / 4 ) S 1 , etc. Look at ln ( S n ) , and use ln ( 1 + t ) t , t > 0 . )

Continuing the hint gives S N = ( 1 + 1 / N 2 ) ( 1 + 1 / ( N 1 ) 2 ( 1 + 1 / 4 ) ) . Then ln ( S N ) = ln ( 1 + 1 / N 2 ) + ln ( 1 + 1 / ( N 1 ) 2 ) + + ln ( 1 + 1 / 4 ) . Since ln ( 1 + t ) is bounded by a constant times t , when 0 < t < 1 one has ln ( S N ) C n = 1 N 1 n 2 , which converges by comparison to the p -series for p = 2 .

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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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