Simple Math: Simplifying (1-1/n) Sequences

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Homework Help Overview

The discussion revolves around simplifying the product of the sequence (1-1/2)(1-1/3)(1-1/4)...(1-1/n), exploring its behavior as n approaches infinity. The subject area includes sequences and mathematical simplification.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants suggest adding fractions within the parentheses to identify patterns. Some express confusion over the implications of extending the sequence to infinity and question the relationship between (n-1)/n and 1-(1/n). Others consider the role of factorials in the simplification process.

Discussion Status

There is an ongoing exploration of the sequence's simplification, with some participants recognizing patterns while others seek clarification on specific mathematical properties. Guidance has been offered regarding the cancellation of intermediate factors, but no consensus has been reached on the overall simplification approach.

Contextual Notes

Some participants mention confusion due to the late hour, which may affect their reasoning. The original post does not specify the extension of the sequence to infinity, leading to varied interpretations among participants.

DivineNathicana
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Greetings. Alright, if anyone's bored enough to be on-line right now, what is the following simplified and how do you get it?

(1-1/2)(1-1/3)(1-1/4)(1-1/5)...(1-1/n)

Thanks for any help,

- Alisa
 
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Add the fractions inside each set of parentheses and see if there is a pattern! :-)
 
I got infinity(1-2/n+2), where n= the denominator of the first of the two fractions being multiplied. That doesn't sound too solid...
 
[tex]\left(1 - \frac {1}{2}\right) \left(1 - \frac {1}{3}\right) \left(1 - \frac {1}{4}\right) \cdot \cdot \cdot \left(1 - \frac {1}{n}\right) = \frac {1}{2} \cdot \frac {2}{3} \cdot \frac {3}{4} \cdot \cdot \cdot \frac {n-1}{n}[/tex]

Do you see a pattern yet?
 
I see the pattern, but I still keep on getting weird-looking answers such as

∞!
---------
((∞-1)!+1)

The (----) being a division sign. If the symbol doesn't come out, it's supposed to be infinity.
 
Well, first off, your original post said nothing about extending it to infinity. But since that seems to be where you are headed consider that

[tex]\frac {n-1}{n} = 1 - \frac {1}{n}[/tex]

Now let n go to infinity! :-)
 
Wait up, it's 2 A.M., and I can't think very straight. Why does (n-1)/n=1-(1/n)? And shouldn't we be doing factorials like ((n-1)!)/n! or something like that since all of this has to be multiplied?
 
DivineNathicana said:
Wait up, it's 2 A.M., and I can't think very straight. Why does (n-1)/n=1-(1/n)? And shouldn't we be doing factorials like ((n-1)!)/n! or something like that since all of this has to be multiplied?

Um ... it's a fundamental property of numbers? The distributive property.

You can certainly use factorials but why would you want to when all the intermediate factors cancel out?
 
Ooh sorry haha I didn't realize what you were talking about. Okay, yeah, so (n-1)/n= 1-(1/n), I see that. So then wouldn't it be just 1/n if we consider all the factoring out?
 
  • #10
Exactly! I knew you'd see it sooner or later. :-)
 
  • #11
Haha thank you! Maybe next time I should try to get started a bit earlier...
 

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