Fibonacci summation

$S = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{16} + \dfrac{5}{32} + \ldots - (1)$

$\dfrac{1}{2}S = 0 + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{2}{16} + \dfrac{3}{32} + \ldots - (2)$

Subtracting $(2)$ from $(1)$ ,

$\dfrac{1}{2} S = \dfrac{1}{2} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{2}{32} + \dfrac{3}{64} + \ldots$

$\Rightarrow \dfrac{1}{2} S = \dfrac{1}{2} + \dfrac{1}{4}S$

$\Rightarrow \dfrac{1}{4} S = \dfrac{1}{2} \Rightarrow S = \boxed{2}$

Required answer is : $-0.25 + 2 = \color{#D61F06}{\boxed{1.75}}$

Let $B=\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{5}{32}+...$ *Notice that $\frac{1}{4}$ is included. Consider the closed form of the Fibonacci Sequence $F_n=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n-\frac{1}{\sqrt{5}}(\frac{1-\sqrt{5}}{2})^n$ such that each term of the series $B$ is given by: $B_n=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{4})^n-\frac{1}{\sqrt{5}}(\frac{1-\sqrt{5}}{4})^n$ It is easy to see that the series is actually two geometric series with $r_1=(\frac{1+\sqrt{5}}{4})$ and $r_2=(\frac{1-\sqrt{5}}{4})$ and that $B=2$ . Since $\frac{1}{4}$ is included in $B$ , the answer to the question is $B-\frac{1}{4}$ or $1.75$

This series can be written as $\large \frac{1}{2} + \frac{1}{4} + \sum _{n=3}^{\infty} \frac{{F}_{n}}{{2}^{n}}$ .

With the Fibonacci sequence, it is known that ${F}_{n} = {F}_{n-1} + {F}_{n-2}$ for $n \ge 3$ with ${F}_{1} = 1$ and ${F}_{2} = 1$ .

Thus, $\large \sum _{n=3}^{\infty} \frac{{F}_{n}}{{2}^{n}}$ can be written as $\large \sum _{n=3}^{\infty} \frac{{F}_{n -1} + {F}_{n - 2}}{{2}^{n}}$

which is equal to $\large \sum _{n=3}^{\infty} \frac{{F}_{n -1}}{{2}^{n}} + \sum _{n=3}^{\infty} \frac{{F}_{n - 2}}{{2}^{n}}$ .

Note that the general terms of those series can be put into the form of $\large \frac{{F}_{n}}{{2}^{n}}$ by shifting indices:

$\large \sum _{n=3}^{\infty} \frac{{F}_{n -1}}{{2}^{n}} + \sum _{n=3}^{\infty} \frac{{F}_{n - 2}}{{2}^{n}} = \frac{1}{2} \sum _{n=2}^{\infty} \frac{{F}_{n}}{{2}^{n}} + \frac{1}{4} \sum _{n=1}^{ \infty} \frac{{F}_{n}}{{2}^{n}}$ .

Now these series looks very similar to the original. By writing out terms from the series, we can reduce both the series to the form of the original:

$\large \frac{1}{2} \sum _{n=2}^{\infty} \frac{{F}_{n}}{{2}^{n}} + \frac{1}{4} \sum _{n=1}^{ \infty} \frac{{F}_{n}}{{2}^{n}} = \frac{{F}_{1}}{8} + \frac{3{F}_{2}}{16} + \frac{3}{4} \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}}$ .

$\large \Longrightarrow \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}} = \frac{{F}_{1}}{8} + \frac{3{F}_{2}}{16} + \frac{3}{4} \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}}$

$\large \Longrightarrow \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}} = \frac{5}{16} + \frac{3}{4} \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}} \Longrightarrow \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}} = \frac{5}{4}$ .

This is true due to the convergence of $\large \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}}$ which can be verified by use of the ratio test which simplifies down to $\large \frac{{F}_{n+1}}{2{F}_{n}}$ which converges to a value less than one as n approaches infinity.

Thus, the answer of $\large \frac{1}{2} + \frac{1}{4} + \sum _{n=3}^{ \infty} \frac{{F}_{n}}{{2}^{n}} = \frac{1}{2} + \frac{1}{4} + \frac{5}{4} = \frac{8}{4} = \boxed{2}$ .