Evaluating this sum involving Fibonacci numbers: $\frac11 + \frac14+\frac2{16}+\frac3{64}+\frac5{256}+\frac8{1024}+\frac{13}{4096}+\cdots$

Question

This question was given to me today on a sheet of questions containing 'sum to infinity' questions. There are no hints as to how to solve it. The series is:

$$\frac11 + \frac14+\frac2{16}+\frac3{64}+\frac5{256}+\frac8{1024}+\frac{13}{4096}+\cdots$$

How can I solve it?

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I wrote the following piece of code to empirically find the answer:

n1 = 0
n2 = 1
total = 0
for i in range(1, 100):
   nth = n1 + n2
   n1 = n2
   n2 = nth
total += (nth/(4**i))
total += 1
print(total)

The program returns the answer $\frac{16}{11}$, this doesn't match the findings of some users.

Answer 1

The closed formula for Fibonacci numbers is in fact completely unnecessary.

Hint 1: Show that the series converges; terms are non-negative and the $k$-th term is less than $1/2^k$ for $k > 1$.

Hint 2: Let $S$ be the sum of the series (which is why we need to show convergence first). Observe that:

  $S+S/4 = \frac11+\frac14+\frac2{16}+\frac3{64}+\frac5{256}+\cdots \\ \quad\quad\quad\quad + \frac14·\left( \frac11+\frac14+\frac2{16}+\frac3{64}+\frac5{256}+\cdots\right)$

  $\quad\quad\quad\quad = \frac11+\left( \frac?4+\frac?{16}+\frac?{64}+\frac?{256}+\cdots \right)$

  $\quad\quad\quad\quad = \frac11+({?}·S-{?})$.

Now find $S$. (Note that we are using the fact that $\sum_{k=1}^∞ f(k) + \sum_{k=1}^∞ g(k) = \sum_{k=1}^∞ (f(k)+g(k))$ for any real functions $f, g$ on $ℕ$ such that both $\sum_{k=1}^∞ f(k)$ and $\sum_{k=1}^∞ g(k)$ converge.)

Answer 2

We recognize in the numerators the Fibonacci sequence

$$F_0=0, F_1=1, F_2=1, F_3=2, \cdots$$.

Therefore, the given series can be written

$$S(x):=\sum_{n=1}^{\infty}F_nx^{n-1} \ \ \text{for} \ \ x=\frac14$$

To be compared with the classical generating function of the $F_n$:

$$\sum_{n=0}^{\infty}F_nx^n=\dfrac{x}{1-x-x^2}$$

implying that

$$S(x)=\dfrac{1}{1-x-x^2}$$

(multiplying or dividing a generating function by the variable $x$ means a shifting operation on the indices of its coefficients) giving

$$S(\tfrac14)=\dfrac{1}{1-\tfrac14-\tfrac{1}{16}}=\dfrac{16}{11}$$

Answer 3

$F_{n}=\frac{1}{\sqrt{5}}(\phi^n-\bar{\phi^n})$, so $\sum\frac{F_{n}}{4^{n-1}}=4\sum\frac{F_{n}}{4^{n}}=\frac{4}{\sqrt{5}}\sum\frac{\phi^n-\bar{\phi^n}}{4^n}$.\ Okey so $\sum\frac{\phi^n}{4^n}=\frac{1}{1-\frac{\phi}{4}}=\frac{4}{4-\phi}=\frac{4}{\frac{8-1-\sqrt{5}}{2}}=\frac{8}{7-\sqrt{5}}$. The same process for $\sum\frac{\bar{\phi^n}}{4^n}$, gives us that $\sum\frac{\bar{\phi^n}}{4^n}=\frac{8}{7+\sqrt{5}}$. So $4\sum\frac{F_{n}}{4^n}=\frac{4}{\sqrt{5}}\sum\frac{\phi^n-\bar{\phi^n}}{4^n}=\frac{4}{\sqrt{5}}(\frac{8}{7-\sqrt{5}}-\frac{8}{7+\sqrt{5}})=\frac{32}{\sqrt{5}}\frac{2\sqrt{5}}{44}=\frac{16}{11}$