Hints and solutions to
Try a Simpler Version #3

Find the sum:

1/(1x2) + 1/(2x3) + 1/(3x4) + ..... + 1/(98x99) + 1/(99x100) = ?

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• This is very similar to yesterday's problem, except this time we have a sum rather than a product to deal with. But we'll approach it the same way, and try the first few in order to see a pattern:
  • 1/(1x2) = 1/2
  • 1/(1x2) + 1/(2x3) = 1/2 + 1/6 = 3/6 + 1/6 = 4/6 = 2/3
  • 1/(1x2) + 1/(2x3) + 1/(3x4) = 2/3 + 1/12 = 8/12 + 1/12 = 9/12 = 3/4
  • 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) = 3/4 + 1/20 = 15/20 + 1/20 = 16/20 = 4/5
  • and so on...
  In doing the calculations above, we took advantage of the fact that to do a given sum, we had already done the previous one and could use the result of the previous calculation.
• There is certainly a pattern in the answers so far! And from the pattern we can infer that the answer to our problem is (probably) 99/100.
• To understand better what is going on, we need to re-express the terms in the sum as follows:
  • 1/(1x2) = 1/2 = 1/1 - 1/2
  • 1/(2x3) = 1/6 = 1/2 - 1/3
  • 1/(3x4) = 1/12 = 1/3 - 1/4
  • and so on...
  This pattern always works, since 1/n - 1/(n+1) = (n+1)/(n(n+1)) - n/(n(n+1)) = 1/(n(n+1)).
• And we can use this pattern to rewrite our sum as follows:
  1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + ... = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) +...

  The first 1 in the 1/1 - 1/2 will always be there, then there will be a lot of cancellation, and then the very last thing subtracted will "survive".

• This shows that the sum up to 1/(n(n+1)) is 1 - 1/(n+1). So our sum really is 1 - 1/100 = 99/100.

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