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Sum of a series question

  • Thread starter zeion
  • Start date
  • #1
467
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Homework Statement



Find the sum of the series.

[tex] \sum_{k=0}^\infty \frac{1}{(k+1)(k+3)} [/tex]


Homework Equations





The Attempt at a Solution


[tex]
= \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \frac{1}{3\cdot5} + ... + \frac{1}{(n+1)\cdot(n+3)}
[/tex]

[tex]
= \frac{1}{2} [(1-\frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + ... + (\frac{1}{(n+1)} - \frac{1}{(n+3)})
[/tex]

[tex]
= \frac{1}{2}[ 1 + (\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) ]
[/tex]

So here
[tex]
(\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) \to 1
[/tex]

[tex]
(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) \to 0
[/tex]

Then the whole thing sums to 1?
 

Answers and Replies

  • #2
402
1
Ok, things went very bad from the second to the third line in your argument; why don't you try to cancel the terms with opposite signs, instead of grouping them?
 

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