Find the series: $\frac{-1}{4}+\left(\frac12+\frac14+\frac28+\frac3{16}+\frac5{32}+\cdots\right)$ Find the series: $$\frac{-1}{4}+\left(\frac12+\frac14+\frac28+\frac3{16}+\frac5{32}+\cdots\right)$$
Evidently, this is a Fibonacci Sequence with a Geometric Sequence. 
But I don't think there is a formula for the sum of Fibonacci Sequence..
Also, I have heard about Binet's formula, but we can't use these formulas. We just have to use elementary, basic things like $S_n-\frac12S_n$ and all. I tried: 
$$S_n=\left(\frac12+\frac14+\frac18+\cdots\right)+\left(\frac18+\frac2{16}+\cdots\right)$$ 
$$S_n=1+\left(\frac18+\frac2{16}+\cdots\right)$$ 
I keep on repeating that, but what happens is that the Fibonacci Sequence gets shifted till infinity. So what to do please tell? 
 A: $$S=\frac{F_1}{2^1}+\frac{F_2}{2^2}+\cdots$$
We want to use the property $F_n+F_{n+1}=F_{n+2}$. Add the sum to itself in such a way that you can use a common denominator:
$$S+\frac{S}{2}=\frac{F_1}{2^1}+\frac{F_1+F_2}{2^2}+\frac{F_2+F_3}{2^3}+\cdots$$
$$\frac{3}{2}S=\frac{F_1}{2}+\underbrace{\frac{F_3}{2^2}+\frac{F_4}{2^3}+\cdots}_{2(S-F_1/2-F_2/4)}$$
Recognize a part of the original sum (with two missing terms and a factor of 2):
$$\frac{3}{2}S=\frac{F_1}{2}+2\left(S-\frac{F_1}{2}-\frac{F_2}{4}\right)$$
Just express $S$, given $F_1=F_2=1$.
$$\frac32S-2S=\frac{1}2-1-\frac12$$
$$\frac{S}{2}=1$$
$$S=2$$
Now your required series will be $$2-\frac14=\frac74$$
A: Let $F_n$ denote the $n^{th}$ Fibonacci number.  So we have the recursion:
$$F_n=F_{n-1}+F_{n-2}\;\;\;F_0=1=F_1$$
Define a power series $$F(x)=\sum F_n\,x^n=1+x +2x^2+3x^3+5x^4+...$$
We want a simple expression for $F(x)$.  But use the recursion and match coefficients of $x^n$...we have $$F(x)=xF(x)+x^2F(x)+1\;\;\Rightarrow\;\;F(x)=\frac{1}{1-x-x^2}$$
Note:  Check this! Easy to drop a term.
Now compare $F(\frac 12)$ to your desired sum.
A: You have the generating function of the Fibonacci sequence:
$$
F_{n + 2} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
$$
$\begin{align}
F(z)
  &= \sum_{n \ge 0} F_n z^n \\
  &= \frac{z}{1 - z - z^2}
\end{align}$
This series converges for $\lvert z \rvert < (\sqrt{5} - 1)/2 \approx 0.618$.
Your sum is essentially $F(1/2)$.
