Proving Summation of \frac{1}{i(i+1)} = \frac{n}{n+1}

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Discussion Overview

The discussion revolves around proving the summation formula \(\sum_{i=1}^n\;\frac{1}{i(i+1)}=\frac{n}{n+1}\). Participants explore various approaches to derive this result, including the concept of telescoping series and manipulation of summations.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant expresses difficulty in deriving the formula after rewriting the summation as two separate sums.
  • Another participant humorously advises against drinking while attempting to derive mathematical results.
  • A participant identifies the series as a telescoping series, noting that most terms cancel out, leaving only a few terms.
  • One participant provides a detailed breakdown of combining the sums and manipulating them to show they equal the original summation.
  • A later reply summarizes the telescoping result, indicating that it simplifies to \(1 - \frac{1}{n+1} = \frac{n}{n+1}\).

Areas of Agreement / Disagreement

Participants generally agree on the telescoping nature of the series and the approach to combine the sums, but there is no consensus on the clarity of the derivation process, as one participant expresses confusion.

Contextual Notes

Some assumptions about the manipulation of summations and the properties of telescoping series may not be explicitly stated, leading to potential gaps in understanding for those unfamiliar with the concepts.

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.
 
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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?
 
It is called a telescoping series. The idea is only one term from each sum remain after all others cancel.
 
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.
 
Telescoping result. 1 - 1/(n+1) = n/(n+1)
 

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