How to prove this series result:
$$\sum_{n=1}^{\infty }\frac{F_{n}}{2^{n}}=2$$
where $F_{1}=1,~F_{2}=1,~F_n=F_{n-1}+F_{n-2},~~n\geq 3$.
I have no idea where to start.
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Olivier Oloa
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http://math.stackexchange.com/questions/137063/prove-that-sum-limits-n-0-infty-fracf-n2n-sum-limits-n-0-i – Jan 31 '16 at 18:19
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1See also http://math.stackexchange.com/questions/114800/infinite-series-fibonacci-2n and http://math.stackexchange.com/questions/88529/how-to-prove-the-fibonacci-sum-sum-limits-n-0-infty-fracf-npn-fr – Martin Sleziak Feb 15 '16 at 16:59
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Hint. You may use the fact that, if
$$f(x)=\sum_{n=0}^\infty F_nx^n \tag1$$
with $F_n$ the nth Fibonnacci number, $F_{n+2}=F_{n}+F_{n+1}$, then
$$f(x)={x\over 1-x-x^2}.\tag2$$
A proof of $(2)$ may be found here. Apply it to $x:=\dfrac12$.
Olivier Oloa
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1A less efficient way is to use $F_n=(a^n-b^n)/\sqrt 5$ where $a=(1+\sqrt 5)/2$ and $b=(1-\sqrt 5)/2=-1/a$. The values $a,b$ are the solutions of $x^2-x-1=0.$ This formula for $F_n$ is easily proved by induction on $n$, noting that $a^{n+1}=a^n+a^{n-1}$ and $b^{n+1}=b^n+b^{n-1}.$ – DanielWainfleet Jan 31 '16 at 08:52