This Proof is right about sums and limits?

[MAGNIBORO]

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
2 image:https://gyazo.com/c6b99485e6d0271c1c0bbdbaaca29d54

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

thanks.

 

[micromass]

No, you have a divergent series.

 

[davidmoore63@y]

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.

 

[MAGNIBORO]

No, you have a divergent series.

I kept working and I think it gets to the real proof
https://gyazo.com/aeee169696eebe6c4357520f9dbaa837

 

[micromass]

You cannot use values like $\infty$ or $\infty-2$ like this. Ever.

 

[MAGNIBORO]

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

 

[MAGNIBORO]

You cannot use values like $\infty$ or $\infty-2$ like this. Ever.

okay and this?
https://gyazo.com/39a967c89bee8109366e6f62991acf26

 

[pwsnafu]

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
$0 = x^{-n} - 2 +x$

 

[MAGNIBORO]

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