This Proof is right about sums and limits?

  • Thread starter MAGNIBORO
  • Start date
  • #1
107
26
hello, sorry for bad English, i have a question.
if we consider the following equations and we take natural values note that tend 2
x-1=0 -----------------> x = 1
x^2-x-1=0 -----------------> x = 1.618033988 (golden ratio)
x^3-x^2-x-1=0 -----------------> x = 1.839286755
x^4-x^3-x^2-x-1=0 -----------------> x = 1.927561975
x^5-x^4-x^3-x^2-x-1=0 -----------------> x = 1.965948236
x^6-x^5-x^4-x^3-x^2-x-1=0 -----------------> x = 1.983582843

then we can assume that when the equation has infinite terms the answer is 2.
and reaches the following "proof" and let me know if it's right:
1 image:https://gyazo.com/49a46e56fb19b4ec7aa21594e4e78cd1 [Broken]
2 image:https://gyazo.com/c6b99485e6d0271c1c0bbdbaaca29d54 [Broken]

Besides knowing if this is OK too I wonder if this is what is called "inductive method"

thanks.
 
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Answers and Replies

  • #2
22,095
3,285
No, you have a divergent series.
 
  • #3
I don't like your proof! If you have
x^n = 1 + x+ x^2 +...+x^(n-1)
using the formula for the sum of a GP you get
x^n = (1-x^n)/(1-x), and so
x - 2 + 1/x^n =0
now you can see that as n tends to infinity x gets as close as you like to 2.
 
  • #4
107
26
No, you have a divergent series.
I kept working and I think it gets to the real proof
https://gyazo.com/aeee169696eebe6c4357520f9dbaa837 [Broken]
 
Last edited by a moderator:
  • #5
22,095
3,285
You cannot use values like ##\infty## or ##\infty-2## like this. Ever.
 
  • #6
107
26
I don't like your proof! If you have
x^n = 1 + x+ x^2 +...+x^(n-1)
using the formula for the sum of a GP you get
x^n = (1-x^n)/(1-x), and so
x - 2 + 1/x^n =0
now you can see that as n tends to infinity x gets as close as you like to 2.
I have a question.
How did you get in here
x^n = (1-x^n)/(1-x)
to here
x - 2 + 1/x^n =0
my poor mind can not compute
 
  • #7
107
26
You cannot use values like ##\infty## or ##\infty-2## like this. Ever.
okay and this?
https://gyazo.com/39a967c89bee8109366e6f62991acf26 [Broken]
 
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  • #8
pwsnafu
Science Advisor
1,080
85
I have a question.
How did you get in here
x^n = (1-x^n)/(1-x)
to here
x - 2 + 1/x^n =0
my poor mind can not compute
Divide both sides by ##x^n## to get
## 1= \frac{x^{-n} - 1}{1-x}##
Multiply both sides by 1-x
##1 - x = x^{-n} -1##
Move everything to left
##0 = x^{-n} - 2 +x ##
 
  • #9
107
26
Divide both sides by ##x^n## to get
## 1= \frac{x^{-n} - 1}{1-x}##
Multiply both sides by 1-x
##1 - x = x^{-n} -1##
Move everything to left
##0 = x^{-n} - 2 +x ##
thank you very much
 

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