Fun with Fibonacci

In order to find this sum let's consider this: $\frac { { F }_{ n+3 } }{ { 2 }^{ n } }$ .

If we try to find the difference of it, we'll get

$\Delta (\frac { { F }_{ n+3 } }{ { 2 }^{ n } } )=\frac { { F }_{ n+3 } }{ { 2 }^{ n } } -\frac { { F }_{ n+2 } }{ { 2 }^{ n-1 } } =\frac { 1 }{ { 2 }^{ n } } ({ F }_{ n+3 }-2{ F }_{ n+2 })=\frac { 1 }{ { 2 }^{ n } } ({ F }_{ n+1 }-{ F }_{ n+2 })=-\frac { { F }_{ n } }{ { 2 }^{ n } }$ .

It means that

$\frac { { F }_{ n } }{ { 2 }^{ n } } =-\Delta (\frac { { F }_{ n+3 } }{ { 2 }^{ n } } )$

And so

$\sum { \frac { { F }_{ n } }{ { 2 }^{ n } } } =-\frac { { F }_{ n+3 } }{ { 2 }^{ n } } +C$ .

Quite easily we can check (placing $n=1$ ) that $C=2$ , which means that

$\sum { \frac { { F }_{ n } }{ { 2 }^{ n } } } =2-\frac { { F }_{ n+3 } }{ { 2 }^{ n } }$ .

And because ${ lim }_{ n\rightarrow \infty }\frac { { F }_{ n+3 } }{ { 2 }^{ n } } =0$ ,

$\sum _{ n=1 }^{ \infty }{ \frac { { F }_{ n } }{ { 2 }^{ n } } } =2$

$\because S=\sum _{ n=1 }^{ \infty }{ \frac { { F }_{ n } }{ { 2 }^{ n } } }\ ------(1)$ We have, $F_1$ = $F_2$ = $1$

$\underline { For\ F_3 }$ : $\because { F }_{ n }={ F }_{ n-1 }+{ F }_{ n-2 }$ $\Rightarrow { F }_{ 3 }={ F }_{ 3-1 }+{ F }_{ 3-2 }$ $\Rightarrow { F }_{ 3 }={ F }_{ 2 }+{ F }_{ 1 }$ $\Rightarrow { F }_{ 3 }=1+1$ $\therefore { F }_{ 3 }=2$ Similarly: ${ F }_{ 4 }=3$ ${ F }_{ 5 }=5$ ${ F }_{ 6 }=8$ $.$ $.$ $.$ $.$ $\infty.$ Now, by putting on the above $F's$ in equation (1) , we get: $S=\sum _{ n=1 }^{ \infty }{ \frac { { F }_{ n } }{ { 2 }^{ n } } }=\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 3 }{ 16 } +\frac { 5 }{ 32 } +\frac { 1 }{ 8 } +.......\infty$ Let's try adding up the first few terms and see what happens. If we add up the first two terms, we get: $\frac { 1 }{ 2 } +\frac { 1 }{ 4 } =\frac { 3 }{ 4 }$ The sum of the first three terms is: $\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =1$ The sum of the first four terms is: $\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 3 }{ 16 } =\frac { 19 }{ 16 }$ And so on. If we write down the partial sums together, i.e. $(\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } ,1,\frac { 19 }{ 16 } ,\frac { 43 }{ 32 } ,.....)$ we can see that the partial sums here form a sequence that has limit $2$ . $\therefore \boxed { S=\sum _{ n=1 }^{ \infty }{ \frac { { F }_{ n } }{ { 2 }^{ n } } }\ =2 }$

This problem is exactly like this one. A Fibonacci Sum

Define $a_n=\frac{F_n}{2^n}$ . Then utilizing the properties of Fibonacci number, we readily have for $a_n=\frac{1}{2}a_{n-1}+\frac{1}{4}a_{n-2}, n\geq 3$ . Sum both sides up to infinity and let the required limit be $S$ (which can be argued to be finite). We have, $S-(a_1+a_2)=\frac{1}{2}(S-a_1)+\frac{1}{4}S$ . Solving for $S$ , we get $S=2$ .