Theorem: Sum of nth Fibonacci Number over nth Power of 2

:$\ds \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n} = 2$
where $F_n$ is the $n$th Fibonacci number.

Proof:
First, let:
:$S = \ds \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n}$
Some sundry results:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n F_k
      | r = F_{n + 2} - 1
      | c = Sum of Sequence of Fibonacci Numbers
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = F_{n + 2}
      | r = \paren {\sum_{k \mathop = 0}^n F_k} + 1
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \frac 1 {2^n}
      | r = \frac 1 {1 - \frac 1 2}
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \frac 1 {1 / 2}
      | c = 
}}
{{eqn | n = 2
      | r = 2
      | c = 
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = S
      | r = \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n}
      | c = 
}}
{{eqn | r = \frac {F_0} {2^0} + \frac {F_1} {2^1} + \sum_{n \mathop = 2}^\infty \frac {F_n} {2^n}
      | c = 
}}
{{eqn | r = \frac 0 1 + \frac 1 2 + \sum_{n \mathop = 2}^\infty \frac {F_n} {2^n}
      | c = {{Defof|Fibonacci Numbers}}: $F_0 = 0$ and $F_1 = 1$
}}
{{eqn | r = \frac 1 2 + \sum_{n \mathop = 2}^\infty \frac {F_n} {2^n}
      | c = 
}}
{{eqn | r = \frac 1 2 + \sum_{n \mathop = 0}^\infty \frac{F_{n + 2} } {2^{n + 2} }
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = \frac 1 2 + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \frac {F_{n + 2} } {2^n} }
      | c = as $2^{n + 2} = 2^2 \times 2^n = 4 \times 2^n$
}}
{{eqn | r = \frac 1 2 + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \frac 1 {2^n} \paren {\paren {\sum_{k \mathop = 0}^n F_k} + 1} }
      | c = from $(1)$
}}
{{eqn | r = \frac 1 2 + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac {F_k} {2^n} } + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \frac 1 {2^n} }
      | c = 
}}
{{eqn | r = \frac 1 2 + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac {F_k} {2^n} } + \frac 2 4
      | c = from $(2)$
}}
{{eqn | n = 3
      | r = 1 + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac {F_k} {2^n} }
      | c = 
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = 1 + \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac {F_k} {2^n} }
      | r = S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 1 4 \paren {\sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac{F_k} {2^n} }
      | o = <
      | r = S
      | c = 
}}
{{end-eqn}}
Hence we can show that $S$ is absolutely convergent by the Ratio Test:
For $S$ we get $a_n = \dfrac {F_n} {2^n}$ therefore
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n } }
      | r = \lim_{n \mathop \to \infty} \size {\frac {\frac {F_{n + 1 } } {2^{n + 1 } } } {\frac {F_n } {2^n } } }
      | c = substituting $a_n$ and $a_{n + 1}$ from above
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {2^n F_{n + 1 } } {2^{n + 1 } F_n} }
      | c = reformatting division of fractions
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {F_{n + 1} } {2F_n} }
      | c = cancelling $2^n$ from top and bottom
}}
{{eqn | r = \lim_{n \rightarrow \infty} \size {\frac {F_n + F_{n - 1} } {F_n + F_n} }
      | c = {{Defof|Fibonacci Numbers}}: $F_{n + 1} = F_n + F_{n - 1}$ and $2F_n = F_n + F_n$
}}
{{eqn | o = <
      | r = 1
      | c = $F_n > F_{n - 1}$ so $F_n + F_n > F_n + F_{n - 1}$
}}
{{end-eqn}}
Therefore the double summation is absolutely convergent and we can exchange the order of the sums.
We note that:
:$0 \le k \le n < \infty$
Therefore by Fubini's Theorem for Infinite Sums:
:$\ds \sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac {F_k} {2^n} = \sum_{k \mathop = 0}^\infty \sum_{n \mathop = k}^\infty \frac{F_k} {2^n}$
and it follows that:
{{begin-eqn}}
{{eqn | l = S
      | r = 1 + \frac 1 4 \paren {\sum_{k \mathop = 0}^\infty \sum_{n \mathop = k}^\infty \frac {F_k} {2^n} }
      | c = continuing from $(3)$
}}
{{eqn | r = 1 + \frac 1 4 \paren {\sum_{k \mathop = 0}^\infty \paren {F_k \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2^k} } \frac 1 {2^n} } }
      | c = reindexing the sum
}}
{{eqn | r = 1 + \frac 1 4 \paren {\sum_{k \mathop = 0}^\infty \paren {\frac 1 {2^k} F_k \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2^n} } } }
      | c = 
}}
{{eqn | r = 1 + \frac 1 4 \paren {2 \sum_{k \mathop = 0}^\infty \frac {F_k} {2^k} }
      | c = from $(2)$
}}
{{eqn | r = 1 + \frac S 2
      | c = as $S = \ds \sum_{k \mathop = 0}^\infty \frac {F_k} {2^k}$ {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac S 2
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = S
      | r = 2
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}