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The given series is: Let T n be its nth term Let be the sum of n terms of the series: = DEDUCTION: approach to zero as n approach to infinity ∴ - Wired Faculty

Question

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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 T_n be its nth term $straight T subscript straight n space equals space fraction numerator 1 over denominator left square bracket nth space term space of space 1 comma space 2 comma space 3 comma space........ right square bracket space left square bracket nth space term space of comma space 2 comma space 3 comma space 4 comma space.......... right square bracket end fraction$
$equals space fraction numerator 1 over denominator left square bracket 1 plus left parenthesis straight n minus 1 right parenthesis 1 right square bracket space left square bracket 2 plus left parenthesis straight n minus 1 right parenthesis 1 right square bracket end fraction space equals space fraction numerator 1 over denominator straight n left parenthesis 2 plus straight n minus 1 right parenthesis end fraction space equals space fraction numerator 1 over denominator straight n left parenthesis straight n plus 1 right parenthesis end fraction$
$rightwards double arrow$    $space space space space space straight T subscript straight n space equals space 1 over straight n minus fraction numerator 1 over denominator straight n plus 1 end fraction$
Let $space space straight S subscript straight n$ 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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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.