Fibonacci Sum

Let $S(x) = \sum_{n=1}^\infty F_n \times x^n$ , and $T(x) = \sum_{n=1}^ \infty n\cdot F_{n} \times x^n$ . Then, we are interested in $S( \frac{1}{2} )$ and $T ( \frac{1}{2} )$ .

According to Fibonacci series , the sum of the power series can be written as:

$\sum_{n=1}^\infty F_n x^n = \frac{x}{1-x-x^2}.$

Setting $x =\dfrac{1}{2}$ , we obtain

$\sum_{n=1}^\infty \frac{F_n}{2^n} = \frac{\frac{1}{2}}{1-\frac{1}{2}-(\frac{1}{2})^2} = 2.$

In order to evaluate $T(x)$ , we will differentiate $S(x)$ : $\sum_{n=1}^\infty F_n \times n \times x^{n-1} = S'(x)= \frac{d}{dx} \left( \frac{ x}{ 1- x- x^2 } \right) = \frac{1\cdot(1-x-x^2) - (-2x-1)(x)}{(1-x-x^2)^2} = \frac{x^2 +1}{(1-x-x^2)^2}.$

Thus,

$T(x) = \sum_{n=1}^\infty (F_n)(n) x^n = x\cdot S'(x) = \frac{x^3 +x}{(1-x-x^2)^2}.$

Now, we can plug in $x =\dfrac{1}{2}$ to obtain:

$T(\frac{1}{2}) = \frac{(\frac{1}{2})^3 + \frac{1}{2}}{(1-\frac{1}{2}-(\frac{1}{2})^2)^2} = \frac{\frac{5}{8}}{\frac{1}{16}} =10$

Therefore, $T(x) = 5\cdot S(x)$ . As a result, $k = \boxed{5}$ .

Let $\displaystyle a_m = \sum_{n=m}^\infty \frac{F_n}{2^n}$ . The LHS converges absolutely, and so can be rearranged into $\displaystyle \sum_{m=1}^\infty \sum_{n=m}^\infty \frac{F_n}{2^n} = \sum_{m=1}^\infty a_m$ (to see this, note that $\frac{F_n}{2^n}$ appears $n$ times, in $m = 1, 2, \ldots, n$ ), and the RHS is simply $k a_1$ . We will now compute a closed form expression for $a_m$ .

On one hand, $a_{m-1} = a_m + \frac{F_{m-1}}{2^{m-1}}$ or $a_m = a_{m-1} - \frac{F_{m-1}}{2^{m-1}}$ ; we simply take the term $n = m-1$ out of the summation. Applying this for $m \gets m-1$ , we have $a_{m-1} = a_{m-2} - \frac{F_{m-2}}{2^{m-2}}$ . Summing the two together, we also get $a_m = a_{m-2} - \frac{F_{m-1}}{2^{m-1}} - \frac{F_{m-2}}{2^{m-2}}$ .

On the other hand, using the Fibonacci recurrence relation,

$\displaystyle\begin{aligned} a_m &= \sum_{n=m}^\infty \frac{F_n}{2^n} \\ &= \sum_{n=m}^\infty \frac{F_{n-1} + F_{n-2}}{2^n} \\ &= \sum_{n=m}^\infty \frac{F_{n-1}}{2^n} + \sum_{n=m}^\infty \frac{F_{n-2}}{2^n} \\ &= \sum_{n=m-1}^\infty \frac{F_n}{2^{n+1}} + \sum_{n=m-2}^\infty \frac{F_n}{2^{n+2}} \\ &= \frac{1}{2} a_{m-1} + \frac{1}{4} a_{m-2} \end{aligned}$

Using these relations, we can obtain the closed form:

$\displaystyle\begin{aligned} a_m &= a_m \\ a_{m-2} - \frac{F_{m-1}}{2^{m-1}} - \frac{F_{m-2}}{2^{m-2}} &= \frac{1}{2} a_{m-1} + \frac{1}{4} a_{m-2} \\ a_{m-2} - \frac{F_{m-1}}{2^{m-1}} - \frac{F_{m-2}}{2^{m-2}} &= \frac{1}{2} \left( a_{m-2} - \frac{F_{m-2}}{2^{m-2}} \right) + \frac{1}{4} a_{m-2} \\ \frac{1}{4} a_{m-2} &= \frac{F_{m-1}}{2^{m-1}} + \frac{1}{2} \cdot \frac{F_{m-2}}{2^{m-2}} \\ a_{m-2} &= \frac{F_{m-1}}{2^{m-3}} + \frac{F_{m-2}}{2^{m-3}} \\ a_{m-2} &= \frac{F_m}{2^{m-3}} \end{aligned}$

Thus, $a_m = \frac{F_{m+2}}{2^{m-1}}$ . Substituting this to the LHS gives

$\displaystyle\begin{aligned} \sum_{m=1}^\infty a_m &= \sum_{m=1}^\infty \frac{F_{m+2}}{2^{m-1}} \\ &= 8 \sum_{m=1}^\infty \frac{F_{m+2}}{2^{m+2}} \\ &= 8 \sum_{m=3}^\infty \frac{F_m}{2^m} \\ &= 8 a_3 \end{aligned}$

Now we can use our closed form expression to obtain $a_1 = 2, a_3 = \frac{5}{4}$ . Thus the LHS is equal to $8 a_3 = 10$ , and the RHS is equal to $k a_1 = 2k$ , so $k = \frac{10}{2} = \boxed{5}$ .

Note that this is easily generalizable for a different base of the denominator; most 2's are simply replaced by the appropriate base.