Question 1184837
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                    I will answer part  (i),   ONLY.




(i)   1/(1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*6) - ...
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<pre>
{{{1/(1*2)}}} = {{{1}}} - {{{1/2}}}


{{{1/(2*3)}}} = {{{1/2}}} - {{{1/3}}}


{{{1/(3*4)}}} = {{{1/3}}} - {{{1/4}}}


{{{1/(4*5)}}} = {{{1/4}}} - {{{1/5}}}


{{{1/(5*6)}}} = {{{1/5}}} - {{{1/6}}}


  . . . . . . . . . . . . 


{{{1/(k*(k+1))}}} = {{{1/k}}} - {{{1/(k+1)}}}



Now take the alternate sum and get


    S = (1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*6) - . . . = ({{{1}}} - {{{1/2}}}) - ({{{1/2}}} - {{{1/3}}}) + ({{{1/3}}} - {{{1/4}}}) - ({{{1/4}}} - {{{1/5}}}) + ({{{1/5}}} - {{{1/6}}} ) + . . . = 

   
      = 1 - 2/2 + 2/3 - 2/4 + 2/5 - 2/6 + . . . = (2 - 2/2 + 2/3 - 2/4 + 2/5 - 2/6  + . . . ) - 1 = 2*(1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + . . . ) - 1 = 


      = 2*ln(1-(-1)) - 1 = 2*ln(2) - 1 = 0.386294361  (rounded).    <U>ANSWER</U>
</pre>

Solved.


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For more details, &nbsp;see this link    


https://www.quora.com/What-is-1-1-2+1-3-1-4+1-5-1-6