A Lucas Sum

Step 1:

First, we evaluate $p$ . Noticing that $p^{2}-1 = p$ , solving for $p$ and discarding the negative solution gives

$p = \frac{1+\sqrt{5}}{2}$

which is the golden ratio $\phi$ .

Step 2:

Now, we use the fact that $\lfloor \phi^{k} \rceil = \lfloor \phi^{k}+\frac{1}{2} \rfloor = \lfloor \phi^{k}+(-\phi)^{-k} \rfloor$ since $\frac{1}{2} > (-\phi)^{-k}$ when $k > 1$ .

It turns out that $L_{k} = \phi^{k}+(-\phi)^{-k}$ is actually the closed form for an integer sequence called the Lucas numbers $\lbrace L_{k} \rbrace_{k=0}^{\infty}$ whenever $k > 1$ . The Lucas numbers follow the recursive rule

$L_n \equiv \begin{cases} 2 & \text{if} & n = 0; \\ 1 & \text{if} & n = 1; \\ L_{n-1}+L_{n-2} & \text{if} & n > 1. \end{cases}$

Step 3:

Hence, we let the sum be $S$ , and rewrite the sum as

$S = \sum_{k=2}^{\infty} \frac{L_{k}}{2^{k}}$

and proceed with a bit of algebraic manipulation:

$S = \frac{3}{4}+\frac{4}{8}+\frac{7}{16}+\frac{11}{32}+\ldots$

$2S = \frac{3}{2}+\frac{4}{4}+\frac{7}{8}+\frac{11}{16}+\ldots$

$2S-S = \frac{3}{2}+\frac{1}{4}+\frac{3}{8}+\frac{4}{16}+\ldots$

$= \frac{3}{2}+\frac{1}{4}+\frac{S}{2}$

$S = 2(\frac{3}{2}+\frac{1}{4}) = \boxed{\frac{7}{2}}$

Sorry, Mr. Lucas and the title of the problem, I didn't actually use Lucas numbers(because I didn't know about it).

It is clear that,

$p = \frac{1 + \sqrt{5}}{2} = \varphi$

So, we just need to find

$\sum _{k=2}^{\infty} \frac{\lfloor \varphi^{k} \rceil}{2^{k}}$

Now, we know that $\varphi^{k} = F_{k}\varphi + F_{k-1}$ , where $F_{k}$ is the $k^{th}$ fibonacci term.

Hence, the numerator of the term of the sum becomes

$\lfloor F_{k}\varphi + F_{k-1} \rceil \Rightarrow \lfloor F_{k} \varphi \rceil + F_{k-1}$

Hence, this becomes

$\sum _{k=2}^{\infty} \frac{\lfloor F_{k}\varphi \rceil}{2^{k}} + \frac{F_{k-1}}{2^{k}}$

2nd Sum

$\sum_{k=2}^{\infty} \frac{F_{k-1}}{2^{k}} = S_{1}$

$2S_{1} = \frac{F_{1}}{2} + \frac{F_{2}}{2^{2}} + \ldots$

Subtracting term-wise,

$S_{1} = \frac{F_{1}}{2} + 0 + \frac{F_{1}}{2^{3}} \ldots$

$S_{1} = \frac{1}{2} + \frac{S_{1}}{2} \Rightarrow S_{1} = 1$

1st Sum

$\sum _{k=2}^{\infty} \frac{\lfloor F_{k}\varphi\rceil}{2^{k}}$

Now, by Binet's formula,

$F_{k} = \frac{\varphi^{k} - (\varphi-\sqrt{5})^{k}}{\sqrt{5}}$

which tells us that

$F_{k}\varphi = \frac{\varphi^{k+1} - \varphi(\varphi-\sqrt{5})^{k}}{\sqrt{5}}$

$= \frac{\varphi^{k+1} - (\varphi-\sqrt{5})^{k+1} - \sqrt{5}(\varphi-\sqrt{5})^{k}}{\sqrt{5}}$

$= F_{k+1} - (\varphi - \sqrt{5})^{k}$

Now, $(\varphi - \sqrt{5})^{k}$ is a decreasing function of $k$ with tenth digit place above 5 only for $k \le 1$ but we have $k>1$ in the sum. So, in the nearest integer function, this term will be trivial.

$\lfloor F_{k}\varphi \rceil = F_{k+1}$

Hence, the sum becomes

$\sum _{k=2}^{\infty} \frac{F_{k+1}}{2^{k}}$

which can be found in the same way as done before for sum 1, getting

$\sum _{k=2}^{\infty} \frac{F_{k+1}}{2^{k}} = \frac{5}{2}$

As a result,

$\sum _{k=2}^{\infty} \frac{\lfloor \varphi^{k} \rceil}{2^{k}} = \frac{5}{2} + 1 = \boxed{\frac{7}{2}}$