Unusual series

Believe it or not, not only the Fibonacci numbers produce the golden ratio. $\huge{ALMOST}$ all number sequences defined by $X_n=X_{n-1}+X_{n-2}$ will approach the golden ratio as $\displaystyle \lim_{n\rightarrow \infty} \dfrac{X_n}{X_{n+1}}$ assuming $X_1,X_2>0$

But now for the proof. Assuming that this limit exists as some finite value $P$ , then we can state that.

$X_n=X_{n-1}+X_{n-2}$

$\frac{X_n}{X_{n-1}}=\frac{X_{n-1}}{X_{n-2}}=P$

$\frac{X_{n-1}+X_{n-2}}{X_{n-1}}=\frac{X_{n-1}}{X_{n-2}}=P$

$1+\frac{X_{n-2}}{X_{n-1}}=\frac{X_{n-1}}{X_{n-2}}=P$

$1+\dfrac{1}{P}=P$

$0=P^2-P-1$

$P=\dfrac{1\pm \sqrt5}{2}$

$P=\phi$

$\cfrac{X_{n+1}}{X_n} \\ =\cfrac{X_n+X_{n-1}}{X_n} \\ =1+\cfrac{X_{n-1}}{X_n} \\ =1+\cfrac{1}{\quad \cfrac {X_n}{X_{n-1}} \quad} \\ =1+\cfrac{1}{\quad\cfrac{X_{n-1}+X_{n-2}}{X_{n-1}}}\\ =1+\cfrac{1}{1+\cfrac{X_{n-1}}{X_{n-2}}}\\ =1+\cfrac{1}{1+\cfrac{1}{1+ \ddots}}$

$Let A=1+\cfrac{1}{1+\cfrac{1}{1+ \ddots}}\\ A-1=\cfrac{1}{1+\cfrac{1}{1+ \ddots}}\\ \cfrac{1}{A-1}=1+\cfrac{1}{1+ \ddots}=A\\ A^2-A-1=0\\ A=\cfrac{\sqrt{5}+1}{2}=\phi$

It's easy to show, by induction, that there exist constants $c_1, c_2 \in \mathbb{R}$ such that $X_n = c_1 \phi^n + c_2 \bar{\phi}^n$ for all $n \in \mathbb{N}$ , where $\phi = \frac{1 + \sqrt{5}}{2}$ and $\bar{\phi} = \frac{1 - \sqrt{5}}{2}$ . (This is a well-known theorem; I didn't just come up with it :P). Therefore,

$\lim_{n \to \infty} \frac{X_{n+1}}{X_n} = \lim_{n \to \infty} \frac{ c_1 \phi^{n+1} + c_2 \bar{\phi}^{n+1}}{ c_1 \phi^n + c_2 \bar{\phi}^n}$ $= \lim_{n \to \infty} \frac{\left(c_1 \phi^{n+1} + c_2 \bar{\phi}^{n} \phi\right) + \left( c_2 \bar{\phi}^{n+1} - c_2 \bar{\phi}^{n} \phi\right)}{ c_1 \phi^n + c_2 \bar{\phi}^n}$ $= \phi + \lim_{n \to \infty} \frac{c_2 \bar{\phi}^{n} \left(\bar{\phi} - \phi \right)}{ c_1 \phi^n + c_2 \bar{\phi}^n} = \phi + \lim_{n \to \infty} \frac{-c_2 \bar{\phi}^{n} \sqrt{5}}{ X_n}$

Now let's consider the remaining limit. In the numerator we have a constant times $\bar{\phi}^n$ . Since $-1 < \bar{\phi} < 0$ , this means the numerator approaches zero. Now the denominator is $X_n$ . By induction, we can show that $X_n$ is always a natural number and strictly increasing, and therefore goes to infinity. Hence, the limit is of the form $\frac{0}{\infty}$ , so it equals $0$ .

Therefore,

$\lim_{n \to \infty} \frac{X_{n+1}}{X_n} = \boxed{\phi}$

Leave all the mathematical works alone ,.......Basically if any sequence is formed by adding its previous terms , then the sequence term ratio Converges to the "Golden ratio " irrespective of the starting numbers

https://www.facebook.com/photo.php?fbid=719045918175887&set=a.331274213619728.79067.100002114550242&type=1

Although the value of limit can be computed as ∅

But lets take it another way for the given set of options:

$lim_{n→∞} X_{n+1}/X_{n}$

$=lim_{n→∞} (X_{n}+X_{n-1})/X_{n}$

$=lim_{n→∞} 1+(X_{n-1}/X_{n})$

$=1+(<1)$

$<2$

So out of the options only ∅ satisfies this condition!