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Prove Limit Of Fn/ Fn-1 As It Approches Inginity Equals Golden Ratio

How to get the golden ratio with an infinite series?

The golden ratio can also be defined as
a number when 1 is subtracted equals the reciprocal of that number or:
GR -1 = 1 / (GR)

Then we get the equation
GR^2 - GR -1 = 0

Using the quadratic formula:
GR = (--1 +-sqrt(1-4*1*-1)) / 2* 1

GR = (1 +- sqrt(1 +4)) / 2

GR = 1 +- sqrt (5) / 2

GR = 1.6180339887499...

We can test this number:

1.6180339887499 -1 should equal 1 / 1.6180339887499

1.6180339887499 -1 = (1 / 1.6180339887499)

.6180339887499 = .6180339887499
Okay well, I guess you still want to know how to solve for this using an infinite series.

Fibonacci numbers (as you probably know) are generated by summing the previous 2 numbers such that
1 + 1 = 2
2 + 1 = 3
2 + 3 = 5
5 + 3 = 8
5 + 8 =13
8 + 13 = 21
13 + 21 = 34
21 + 34 = 55
34 + 55 = 89
55 + 89 = 144

"phi" (aka the golden ratio) is equal to a Fibonacci number divided by its previous Fibonacci number.

So, 3/2 = 1.5
5/3 = 1.66666666666666...
8/5 = 1.6
13/8 = 1.625
21/13 = 1.61538461538462
34/21 = 1.61904761904762

We already know that "phi" is approximately equal to 1.61803398874990... and this ratio is reached when we get to the 42nd Fibonacci number 267,914,296 divided by the 41st number 165,580,141. At this point the ratio "stays" at 1.61803398874990... and any further accuracy is limited by my computer. So at least you could definitely state that this value is accurate to 14 decimal places.

How do I prove that a limit exists for the Fibonacci ratio Fn+1 divided by Fn as n approaches infinity?

Using the relationship F(n+1)F(n-1)-{F(n)}^2 = +/-1 (proved below) and that F(n) > n when n>5, write |F(n+1)/F(n) - F(n)/F(n-1)| = 1/{F(n)F(n-1)} <= 1/{n(n-1)} = 1/(n-1) - 1/n. Now using the triangle inequality gets |F(m+1)/F(m) - F(n+1)/F(n)| <= |1/m - 1/n| which is certainly a Cauchy sequence, so F(n+1)/F(n) has a unique limit L. Since F(n+1) = F(n) + F(n-1) and F(n-1) <= F(n) then 1 < F(n+1)/F(n) < 2 so L is finite.

The final step is to find the limit L. Since F(n+1)/F(n) = 1 + 1/(F(n)/F(n-1)), then letting n --> infinity L = 1+ 1/L so L^2-L-1=0 and L is the golden ratio (1+sqrt(5))/2
To show F(n+1)F(n-1) - {F(n)}^2 =+/-1 use induction on n and the relation F(n+1) = F(n) +F(n-1). Writing F(n+1)F(n-1) - F(n)F(n) = (F(n)+F(n-1))F(n-1) - (F(n-1) +F(n-2))F(n) = F(n-1)F(n-1) - F(n)F(n-2) = -(F(n)F(n-2) - F(n-1)F(n-1)) = +/-1 by induction, since F(3)F(1)-F(2)F(2) = 2*1-1*1= 1.

Would you agree that [math]\lim_{n\to\infty}n = \infty[/math]?  For all n past 5, [math]\text{Fib}(n) > n[/math], so the limit of the Fibonacci sequence had better be at least as large as the limit of n.[math]\text{Fib}(n)[/math]: 1, 1, 2, 3, 5, 8, ...[math]n[/math]: 1, 2, 3, 4, 5, 6, ...

Prove that 5 divides fn if and only if 5 divides n?

Is fn the nth fibonacci number?

If so, look at the numbers mod 5.

I guess the fast way to prove this is:

Theorem: For any k>=0, F(5k) is divisible by 5 and
F(5k+1), F(5k+2), F(5k+3), and F(5k+4) are not divisible by 5.


If k=0, then:

F(0) = 0 is divisible by 5.
F(1), F(2), F(3), and F(4) are 1,1,2, and 3, respectively, so it is true for 5*0+r where r=1,...4.

Now we prove for k given that it is true for k-1.

In particular, we know that:

F(5k-3) = F(5(k-1)) + F(5k-4)
F(5k-2) = F(5k-4) + F(5k-3) = F(5(k-1)) + 2F(5k-4)
F(5k-1) = F(5k-3) + F(5k-2) = 2F(5(k-1)) + 3F(5k-4)
F(5k) = F(5k-2) + F(5k-1) = 3F(5(k-1)) + 5F(5k-4)

So since F(5(k-1)) is divisible by 5, and 5F(5k-4) is divisible by 5, then F(5k) is divisible by 5.

F(5k+1) = F(5k) + F(5k-1)

But 5k-1 = 5(k-1)+4, so F(5k-1) is not divisible by 5, so F(5k+1) can't be divisible by 5.

F(5k+2) = F(5k) + F(5k+1)

Again, since F(5k) is divisible by 5 and F(5k+1) is not, F(5k+2) is not.

F(5k+3) = F(5k+1) + F(5k+2) = 2F(5k+1) + F(5k)

Since F(5k+1) is not divisible by 5, then 2F(5k+1) can't be divisible by 5, so F(5k+3) can't be divisible by 5.

F(5k+4) = F(5k+3) + F(5k+2) = 3F(5k+1)+2F(5k)

So again, if F(5k+4) is divisible by 5, then F(5k+1) would be, which we know already is false.

So we have proven it by induction.

Mathematical proof that god exists?

This was the theory I came up with this afternoon that was since disproven

The ratio of the circumference of a circle to its diameter is 3.14
So if the world is a sphere
Then its (4/3)3.14R^3
If hell is at the end of the radius and the Earth is 500 million light years thick
Then the Earth is one billion light years across
If the universe is only ~4billion years old
Then that means that a cubic million light years of matter accumulated here every 8 years
If we go back to when hell was invented by good 2 billion years ago
2^3 is eight, see the relationship between Earth's accumulation and when hell was created?
so we take 8 and substitute it as the radius of the Earth
we multiply it 4 times since its a sphere
we divide it by 3 since 4/3 is 1.33 which is the gradient of the outside of the sphere
then we cube it because its a 3 dimensional amount of matter
then if we round it we put in 0 for god not existing into the equation for radius
then we get 0 which means that the Earth doesn't exist

I attempted to use calculus to find him on a graph. If god exists as 0 and everything exists as infinity and we divide by god, then the graph does not exist. therefore, god has to exist for reality to balance with calculus.

This was also disproven.

I think I am getting close though. What is we tried to find the volume of the ocean, divided it by 666, added the surface area of god which is the surface area of the universe, and then subtracted the golden number. can someone please do this for me.

It’s often easier to solve a more general problem, than a specific one (with 2018s in it).Let’s replace any 2018 in the exponents by [math]n[/math], and any other 2018 by [math]a[/math]:[math]\displaystyle \lim_{x\to\infty}\frac{(x+1)^n+(x+2)^n+\cdots+(x+a)^n}{x^n+a^n}[/math]Now divide both nominator and denominator by [math]x^n[/math]:[math]\displaystyle \lim_{x\to\infty}\frac{(1+\frac1x)^n+(1+\frac2x)^n+\cdots+(1+\frac ax)^n}{1+(\frac ax)^n}[/math]Finally take the limit by taking the limit of individual terms:[math]\displaystyle \frac{1^n+1^n+\cdots+1^n}1[/math]Which clearly equals the number of terms, i.e., [math]a[/math].So the answer is 2018, the number of terms in the nominator. The value of the exponents has nothing to do with it.

Irrational Numbers and Real Life.?

This is a bit of a trick question. You can't use an irrational number in any application. You can only use an approximation. That's the definitional of irrational.

So, the length of anything is irrational. Anytime you measure anything, you are obtaining an approximation to an irrational number. The temperature of your oven, the speed of your car, anything that is continuous is irrational and can only be approximated. It is sort of a use of an irrational number.

For an answer more in line of what was intended. Suppose you want to draw a pleasing rectangle. One that isn't too fat or too skinny. You would make the sides proportional to phi, the golden ratio. Advertisers use this ratio all the time. Phi = 1.618......

MS in math, government mathematician for 38 years.