Answer to Question #273379 in Calculus for 123

Question #273379

If Fn is the n-th Fibonacci number, show that lim "\\lim_{n \\to \\infty} \\frac { F_{n+1}}{F_n} = \\frac{1+\\sqrt{5}}{2}"





1
Expert's answer
2021-11-30T16:06:23-0500

Prove that


"F_n=\\dfrac{1}{\\sqrt{5}}\\bigg((\\dfrac{1+\\sqrt{5}}{2})^{n}-(\\dfrac{1-\\sqrt{5}}{2})^{n}\\bigg)"


Let’s  look at the interesting quadratic


"x^2-x-1=0"

The roots of this quadratic are

"x=\\dfrac{1\\pm\\sqrt{5}}{2}"

This quadratic can also be written as


"x^2=x+1"

From this, we can write expressions for all "x^n:"


"x=x"




"x^2=x+1"

"x^3=x\\cdot x^2=x(x+1)""=x^2+x=x+1+x=2x+1"


"x^4=x\\cdot x^3=x(2x+1)""=2x^2+x=2x+2+x=3x+2"

"x^5=5x+3=x^3+x^4"

"x^6=8x+5=x^4+x^5"

"..."

We note that


"x^n=F_nx+F_{n-1}"

Let the roots of our original quadratic be "u=\\dfrac{1+\\sqrt{5}}{2}," and "v=\\dfrac{1-\\sqrt{5}}{2}."

Since both "u" and "v" are roots of the quadratic, they must both satisfy 

"x^n=F_nx+F_{n-1}"

 So


"u^n=F_nu+F_{n-1}"

"v^n=F_nv+F_{n-1}"

Subtracting the second equation from the first equation yields


"u^n-v^n=F_n(u-v)"

Then


"(\\dfrac{1+\\sqrt{5}}{2})^{n+1}-(\\dfrac{1-\\sqrt{5}}{2})^{n+1}"

"=F_n(\\dfrac{1+\\sqrt{5}}{2}-\\dfrac{1-\\sqrt{5}}{2})"

"=\\sqrt{5}F_n"

Then we have


"F_n=\\dfrac{1}{\\sqrt{5}}\\bigg((\\dfrac{1+\\sqrt{5}}{2})^{n}-(\\dfrac{1-\\sqrt{5}}{2})^{n}\\bigg)"


"F_{n-1}=\\dfrac{1}{\\sqrt{5}}\\bigg((\\dfrac{1+\\sqrt{5}}{2})^{n-1}-(\\dfrac{1-\\sqrt{5}}{2})^{n-1}\\bigg)"

"\\dfrac{F_n}{F_{n-1}}=\\dfrac{(\\dfrac{1+\\sqrt{5}}{2})^{n}-(\\dfrac{1-\\sqrt{5}}{2})^{n}}{(\\dfrac{1+\\sqrt{5}}{2})^{n-1}-(\\dfrac{1-\\sqrt{5}}{2})^{n-1}}"

"=\\dfrac{\\dfrac{1+\\sqrt{5}}{2}-\\dfrac{1-\\sqrt{5}}{2}\\cdot(\\dfrac{1-\\sqrt{5}}{1+\\sqrt{5}})^{n-1}}{1-(\\dfrac{1-\\sqrt{5}}{1+\\sqrt{5}})^{n-1}}"


"\\lim\\limits_{n\\to \\infin}\\dfrac{F_n}{F_{n-1}}="

"=\\lim\\limits_{n\\to \\infin}\\dfrac{\\dfrac{1+\\sqrt{5}}{2}-\\dfrac{1-\\sqrt{5}}{2}\\cdot(\\dfrac{1-\\sqrt{5}}{1+\\sqrt{5}})^{n-1}}{1-(\\dfrac{1-\\sqrt{5}}{1+\\sqrt{5}})^{n-1}}"

"=\\dfrac{\\dfrac{1+\\sqrt{5}}{2}-\\dfrac{1-\\sqrt{5}}{2}\\cdot0}{1-0}"

"=\\dfrac{1+\\sqrt{5}}{2}"


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