Home

Dropping 9-terms from the harmonic series

1,444.0K Views

The harmonic series–that is, 1/1 + 1/2 + 1/3 + 1/4 + …–diverges.  That is, the sum is not finite.  This is in difference to, for example, a geometric series–like ½0 + ½1 + ½2 + ½3 + …–which converges, that is, has a finite sum.

Consider the harmonic series, but drop all terms whose denominator represented in decimal contains a 9.  For example, you’d drop terms like 1/9, 1/19, 1/90, 1/992, 1/529110.  Does the resulting series converge or diverge?

Hint: More generally, you may consider representing the denominator in the base of your choice and dropping terms that contain a certain digit of your choice.
Consider again the harmonic series, but drop a term only if the denominator represented in decimal contains two consecutive 9’s.  For example, you’d drop 1/99, 1/992, 1/299, but not 1/9 or 1/909.  Does this series converge or diverge?

Share
ronret45 Expert Asked on 1st August 2015 in Microsoft Interview Puzzles.
Add Comment
Cancel

    • 1 Answer(s)

    or the first question, the resulting series will converge. To see why, note that the terms that are dropped from the series are of the form 1/(n*10^k + 9), where n is a positive integer and k is a non-negative integer. We can bound each of these terms as follows:

    1/(n10^k + 9) <= 1/(n10^k) = 1/n * 1/(10^k)

    The series of 1/n converges by the integral test, and the series of 1/(10^k) converges to 1/9, since it is a geometric series with common ratio 1/10. Therefore, the series obtained by dropping the terms containing a 9 in the denominator is bounded above by a convergent series, and hence it also converges.

    For the second question, the resulting series will still diverge. To see why, note that the terms that are dropped from the series are of the form 1/(n*10^k + 99), where n is a positive integer and k is a non-negative integer. We can write this as:

    1/(n10^k + 99) = 1/(n10^k) * 1/(1 + 99/(n*10^k))

    As before, the series of 1/(n10^k) converges by the integral test. However, the series of 1/(1 + 99/(n10^k)) diverges, since it is a series of terms that are greater than or equal to 1/100 for sufficiently large n. Therefore, the series obtained by dropping the terms containing two consecutive 9’s in the denominator still contains a divergent subseries, and hence it also diverges.

    Cancel
    Moshe Expert Answered on 27th March 2023.
    Add Comment
    Cancel

    • Your Answer

    By posting your answer, you agree to the privacy policy and terms of service.

  • More puzzles to try-

  • Accenture Interview PuzzlesAdobe Interview PuzzlesAge RiddleAkbar Birbal PuzzlesAlgorithm PuzzlesAlphabet riddleAmazon Interview PuzzlesAnalytical MathematicsAptitude PuzzleBank PuzzlesBetting PuzzlesBrain TeasersCalendar PuzzlesCards PuzzlesChess Board PuzzleChess PuzzlesChristmas Puzzlecipher PuzzleCivil Services PuzzleClock time puzzleCognizant Interview PuzzlesCoins PuzzleComputational PuzzleconundrumCoronavirus PuzzleCoupondunia Interview PuzzleCritical Thinking Puzzledata analyticsData Structure Interview QuestionsDecode PuzzleDetective PuzzlesDice PuzzleDictionary Riddlesdifficult riddleEasy Math puzzlesEasy puzzlesEinstein puzzleEnglish RiddleFamily Tree Puzzlefamous puzzleFill in the blanks riddlesFinding Killer RiddlesFlipkart interview puzzlesFunny RiddlesGeneral KnowledgeGeographical PuzzleGoogle Code Jam 2014Google Interview PuzzlesGRE PuzzleHard Puzzle