Haven't learned calculus yet but I need this for a proof for Fibonacci numbers and its limit In an infinite series of Fibonacci numbers, is this always true $$\frac{F_{n}}{F_{n+1}}>\frac{F_{n-1}}{F_n}$$? Can you make an argument that in an infinite convergent series, eventually that will be false?
 A: hint :We know **$\color{red} {F_{n+1}=F_n+F_{n-1}}\\ \to \color{red} {F_{n}=F_{n+1}-F_{n-1}}$
$$\frac{F_{n+1}-F_{n-1}}{F_{n+1}}>\frac{F_{n-1}}{F_{n+1}-F_{n-1}} $$
or see this $$\begin{bmatrix}f_{n+1} & f_n \\f_n & f_{n-1} \end{bmatrix}=\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}^n \to \det(\begin{bmatrix}f_{n+1} & f_n \\f_n & f_{n-1} \end{bmatrix})=\det(\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}^n)\\f_{n+1}f_{n-1}-f_n.f_n=(-1)^n\\n=2k \to f_{n+1}f_{n-1}-f_n.f_n=+1 \\f_{n+1}f_{n-1}>f_n.f_n\\ \frac{f_{n+1}}{f_{n}}>\frac{f_{n}}{f_{n-1}}\\
n=2k+1 
\to f_{n+1}f_{n-1}-f_n.f_n=-1 \to f_{n+1}f_{n-1}<f_n.f_n \\ \to  \frac{f_{n+1}}{f_{n}}<\frac{f_{n}}{f_{n-1}} \to \\
\color{red} {\frac{f_{n}}{f_{n+1}}>\frac{f_{n-1}}{f_{n}}}$$ this is your case  when $n$ is odd
A: If you look at the fractions $2/3$, $3/5$ and $5/8$ you'll see that the inequality you want isn't always true. In fact it alternates between being true and false. In the limit the sequence (not series) is the golden ratio. Every other term is bigger/smaller; the difference goes to zero.
You can find a proof on the wikipedia page 
https://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio
A: $$\frac{F_{n}}{F_{n+1}}>\frac{F_{n-1}}{F_n}$$
is equivalent to
$$F_{n}^{2} - F_{n-1} F_{n+1} > 0.$$
Using
\begin{align}
5 \, F_{n}^{2} &= L_{2n} - 2 (-1)^{n} \\
5 \, F_{n-1}F_{n+1} &= L_{2n} + 3 (-1)^{n},
\end{align}
where $L_{n}$ are the Lucas numbers, then
$$-5 (-1)^n > 0$$
is the result. Only the odd values of $n$ make the given inequality true, ie
$$\frac{F_{2n+1}}{F_{2n+2}}>\frac{F_{2n}}{F_{2n+1}}$$
