Proving inf(x_n)=1 for x_{n+1}=2-\frac{1}{x_n}

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SUMMARY

The discussion centers on proving that the infimum of the sequence defined by the recurrence relation x_{n+1}=2-\frac{1}{x_n} is equal to 1. Participants emphasize the importance of defining the initial value x_0 to clarify the sequence's behavior. A hint is provided to simplify the proof by subtracting 1 from both sides of the equation, which aids in analyzing the sequence's convergence properties.

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  • Familiarity with the concept of infimum in mathematical analysis
  • Basic knowledge of limits and convergence
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  • Study the properties of recurrence relations in mathematical analysis
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hockey777
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x_{n+1}=2-\frac{1}{x_n}

I need to show that inf(x_n)=1

For someone reason this is proving to be more difficult than I thought, could someone pleas help?
 
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It would be a little clearer if you defined x_0. Also your notation is a little strange - frac{1}{x} means what? I am guessing it means the fractional part, but I am not sure.
 
Welcome to PF!

hockey777 said:
[tex]x_{n+1}=2-\frac{1}{x_n}[/tex]

I need to show that inf(x_n)=1

Hi hockey777! Welcome to PF! :smile:

Hint: start by subtracting 1 from each side. :wink:

(and type [noparse][tex]before and[/tex] after your equations![/noparse] :smile:)
 

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