Sum as formula

  • Thread starter neom
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  • #1
neom
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How can I show that [tex]\sum_{i=1}^n\;\frac1{i(i+1)}=\frac{n}{n+1}[/tex]

I've already figured out i can write it as [tex]\sum_{i=1}^n\;\frac1{i}-\sum_{i=1}^n\;\frac1{i+1}[/tex]

but as I'm a little drunk I can't figure out how to get from there to the formula.

Sorry if I put this in the wrong sextion, but twas in my calculus book.
 

Answers and Replies

  • #2
Mute
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1) Don't drink and derive.

2) What would happen, say, if you wrote the first sum as [itex]1 + \sum_{i=2}^n (1/i)[/itex] and then changed variables in that sum to something convenient?
 
  • #3
lurflurf
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It is called a telescoping series. The idea is only one term from each sum remain after all others cancel.
 
  • #4
HallsofIvy
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If I understand what you are asking, it is just
[tex]\sum_{i= 1}^n\frac{1}{i}+ \sum_{i= 1}^n\frac{1}{i+ 1}= \sum_{i= 1}^n\left(\frac{1}{i}- \frac{1}{i+ 1}\right)[/tex]
[tex]= \sum_{i= 1}^n \left(\frac{i+1}{i(i+1)}- \frac{i}{i(i+1)}\right)= \sum_{i= 1}^n\frac{1}{i(i+1)}[/tex]

In other words, combine the two sums, get common denominators, and subtract the numerators.
 
  • #5
mathman
Science Advisor
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Telescoping result. 1 - 1/(n+1) = n/(n+1)
 

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