Permutations And Combinations

  • Question
    CBSEENMA11013173

    Find the sum to n terms the series fraction numerator 1 over denominator 1.2 end fraction plus fraction numerator 1 over denominator 2.3 end fraction plus fraction numerator 1 over denominator 3.4 end fraction plus........... and deduce the sum to infinity.

    Solution

    The given series is:   space space fraction numerator 1 over denominator 1.2 end fraction plus fraction numerator 1 over denominator 2.3 end fraction plus fraction numerator 1 over denominator 3.4 end fraction plus.......
    Let Tn be its nth term  straight T subscript straight n space equals space fraction numerator 1 over denominator left square bracket nth space term space of space 1 comma space 2 comma space 3 comma space........ right square bracket space left square bracket nth space term space of comma space 2 comma space 3 comma space 4 comma space.......... right square bracket end fraction
                                       equals space fraction numerator 1 over denominator left square bracket 1 plus left parenthesis straight n minus 1 right parenthesis 1 right square bracket space left square bracket 2 plus left parenthesis straight n minus 1 right parenthesis 1 right square bracket end fraction space equals space fraction numerator 1 over denominator straight n left parenthesis 2 plus straight n minus 1 right parenthesis end fraction space equals space fraction numerator 1 over denominator straight n left parenthesis straight n plus 1 right parenthesis end fraction
    rightwards double arrow                        space space space space space straight T subscript straight n space equals space 1 over straight n minus fraction numerator 1 over denominator straight n plus 1 end fraction
    Letspace space straight S subscript straight n be the sum of n terms of the series:
    straight S subscript straight n space equals space sum from straight k equals 1 to straight n of straight T subscript straight k space equals space sum from straight k equals 1 to straight n of open parentheses 1 over straight k minus fraction numerator 1 over denominator straight k plus 1 end fraction close parentheses space equals space open parentheses 1 over 1 minus 1 half close parentheses plus open parentheses 1 half minus 1 third close parentheses plus open parentheses 1 third minus 1 fourth close parentheses plus.... plus open parentheses 1 over straight n minus fraction numerator 1 over denominator straight n plus 1 end fraction close parentheses space equals space 1 minus fraction numerator 1 over denominator straight n plus 1 end fraction 

                                                                                                                                      = fraction numerator straight n plus 1 minus 1 over denominator straight n plus 1 end fraction
    rightwards double arrow             space space space straight S subscript straight n space equals space fraction numerator straight n over denominator straight n plus 1 end fraction
    DEDUCTION:    straight S subscript straight n space equals space space fraction numerator begin display style straight n over straight n end style over denominator begin display style straight n over straight n end style plus begin display style 1 over straight n end style end fraction space equals space fraction numerator 1 over denominator 1 plus begin display style 1 over straight n end style end fraction comma space 1 over straight n  approach to zero as n approach to infinity
    ∴      straight S subscript infinity space equals space fraction numerator 1 over denominator 1 plus 0 end fraction space equals space 1

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