SOLUTION: Please help me answer this question -- Test whether the following infinite series converge, and if they do, find their sum: i) 1/(1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*

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Question 1184837: Please help me answer this question --
Test whether the following infinite series converge, and if they do, find their sum:
i) 1/(1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*6) - ...

ii) 1/(1*2) - 1/(3*4) + 1/(5*6) - 1/(7*8) + 1/(9*10) - ...
Thank you!
Found 2 solutions by ikleyn, robertb:
Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

                    I will answer part  (i),   ONLY.



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


 =  - 


 =  - 


 =  - 


 =  - 


 =  - 


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


 =  - 



Now take the alternate sum and get


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

Solved.

----------------

For more details,  see this link

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




Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
(i) ,

(ii), which I truncated to the first 10 decimal places. THERE IS an exact value for this.

There is a general method which will give the exact values for these series. To get the details, message me, including your email.
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