I am working on a rather long question. I am going to write out the question, and what I've come up with so far. I'm not sure if the question has too many parts to receive answers but I will appreciate any help at all.
The question is
The Fibonacci sequence $(a_{n})_{n=0}^\infty$ is defined by $a_{0}=1$, $a_{1}=1$ and $a_{n+1}:=a_{n}+a_{n-1}$ for all $n\geq 1$.
(a) Show that $\frac {a_{n+1}}{a_{n}}\leq 2$ for $n\geq 0$
(b) Let $f(x):=\sum_{n=0}^{\infty}a_{n}x^{n}$. Show that $f(x)$ converges if $|x|<\frac{1}{2}.$
(c) If $|x|<\frac {1}{2}$ then prove that $f(x)=\frac{1}{1-x-x^{2}}$.
(d) Use the partial fraction decomposition for $\frac {1}{1-x-x^{2}}$ and the geometric series to obtain another power series expression for $f$.
(e) By comparing coefficients, show that $a_{n}=\frac {1}{\sqrt{5}}\left ( \left ( \frac{1+\sqrt{5}}{2} \right )^{n+1}- \left ( \frac{1-\sqrt{5}}{2} \right )^{n+1} \right )$
I will begin with what I've been able to accomplish and my intuition so far, in order of the parts of the question.
(a) I've begun by rearranging as follows:
$\frac {a_{n+1}}{a_{n}}=\frac {a_{n}+a_{n-1}}{a_{n}}\leq 2$ $1+\frac {a_{n-1}}{a_{n}}\leq 2$ $\frac {a_{n-1}}{a_{n}}\leq 1$
So I've got to show that $a_{n-1}\leq a_{n}$ for $n\geq 0$. Intuitively this makes sense to me. The Fibonnaci sequence is constantly increasing (accept for the first two terms). A term will always be smaller or equal to the term that comes after it. I've tried to prove this using induction but I have been unsuccessful. Maybe what I have is sufficient?
(b) This one clearly follows from part (a). I've done the following using the ratio test to find the radius of convergence.
Let $b_{n}=a_{n}x^{n}$ and $b_{n+1}=a_{n+1}x^{n+1}$. So,
$\left | \frac {b_{n+1}}{b_{n}} \right |=\left | \frac {a_{n+1}x^{n+1}}{a_{n}x^{n}} \right | = \frac {a_{n+1}}{a_{n}} \cdot |x|$
Taking the limit
$\lim_{n \to \infty} \frac {a_{n+1}}{a_{n}} \cdot |x|=(\lim_{n \to \infty} \frac {a_{n+1}}{a_{n}}) \cdot |x|=L$
The limit $n \to \infty$ of $\frac {a_{n+1}}{a_{n}}$ can not be greater than 2. I also know that $L$ cannot be greater than 1 in order for there to be convergence. I've tried this
Let $1< \epsilon \leq 2$.
$(\lim_{n \to \infty} \frac {a_{n+1}}{a_{n}}) \cdot |x|=\epsilon |x|<1$
Then $|x|<1/\epsilon$. But this doesn't work. If, for example, as allowed by the conditions I stated, $\epsilon=1.2$, then $|x|>1/2$.
(c) I've done the following based on the Wikipedia entry for Fibonacci sequences.
$f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$
$=a_{0}+a_{1}x+\sum_{n=2}^{\infty}(a_{n-1}+a_{n-2})x^{n}$
$=1+x+\sum_{n=2}^{\infty}a_{n-1}x^{n}+\sum_{n=2}^{\infty}a_{n-2}x^{n}$
$=1+x+x \cdot \sum_{n=0}^{\infty}a_{n}x^{n}+x^{2} \cdot \sum_{n=0}^{\infty}a_{n}x^{n}$
So,
$f(x)=1+x+xf(x)+x^2f(x)=\frac {1+x}{1-x-x^{2}}$
Obviously I've gone wrong somewhere. Taking a different route I recognized that because $|x|<1/2<1$, I can rewrite $f$ as
$f(x)=\frac {1}{1-x} \cdot \sum_{n=0}^{\infty}a_{n}$
This has not proved useful as of yet.
(d) I've found the partial fraction decomposition using Wolfram Alpha to be
$\frac {1}{1-x-x^2}=\frac {2}{(5+\sqrt{5})+(\sqrt{5}\cdot 2x)}+\frac {2}{(5-\sqrt{5})-(\sqrt{5}\cdot 2x)}$
$= \frac {2}{5+\sqrt{5}} \cdot \frac {1}{x \frac{2}{\sqrt{5}+1}+1}+\frac {2}{5-\sqrt{5}} \cdot \frac {1}{x \frac{2}{\sqrt{5}-1}+1}$
$=\frac {2}{5+\sqrt{5}} \cdot \sum_{n=0}^{\infty}(\frac{2x}{\sqrt{5}+1})^{n}+\frac {2}{5-\sqrt{5}} \cdot \sum_{n=0}^{\infty}(\frac{2x}{\sqrt{5}-1})^{n}$