Expert answers to "Prove3π8=∑k≥0(−15)kF2k+12k+1q2k+1where q=2+221+17+825, and Fn are the Fibonacci numbers."

klancimarsn

Answered question

2022-02-23

Prove

\frac{3 π}{8} = \sum_{k \geq 0} {(- \frac{1}{5})}^{k} \frac{F_{2 k + 1}}{2 k + 1} q^{2 k + 1}

where

q = \frac{2 + 2 \sqrt{2}}{1 + \sqrt{\frac{17 + 8 \sqrt{2}}{5}}}

, and

F_{n}

are the Fibonacci numbers.

Answer & Explanation

emeriinb4r

Beginner2022-02-24Added 10 answers

Let
$f (x) = \sum_{k \geq 0} F_{k + 1} x^{k}$
so that
$(1 - x - x^{2}) f (x) = F_{1} + (- F_{1} + F_{2}) x$
or
$f (x) = \frac{1}{1 - x - x^{2}}$
Integrating the above we get
$F (x) = \int_{0}^{x} \frac{d t}{1 - t - t^{2}} = \sum_{k \geq 0} \frac{F_{k + 1}}{k + 1} x^{k + 1}$
The integrand on left side can be written
$\frac{1}{(1 - a t) (1 - b t)} = \frac{A}{1 - a t} + \frac{B}{1 - b t}$
where a, b are roots of $z^{2} - z - 1 = 0$ and
$A = \frac{a}{a - b}, B = - \frac{b}{a - b}$
Letting
$a = \frac{1 + \sqrt{5}}{2}, b = \frac{1 - \sqrt{5}}{2}$
we get $a - b = \sqrt{5}$ and hence we have
$F (x) = \frac{1}{\sqrt{5}} \log \frac{1 - b x}{1 - a x}$
and $G (x) = F (x) - F (- x) = 2 \sum_{k \geq 0} \frac{F_{2 k + 1}}{2 k + 1} x^{2 k + 1}$
Replacing x by ix we get
$\sum_{k \geq 0} {(- 1)}^{k} \frac{F_{2 k + 1}}{2 k + 1} x^{2 k + 1} = - i \frac{G (i x)}{2}$
The sum in question is
$- i \sqrt{5} \frac{G (i \frac{q}{\sqrt{5}})}{2}$
Now
$G (x) = \frac{1}{\sqrt{5}} \log \frac{1 + \sqrt{5} x + x^{2}}{1 - \sqrt{5} x + x^{2}}$
and hence

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