Editing Fibonacci sequence (section)

== Reciprocal sums ==
<!--
Borwein credits some formulae to {{Citation | author = Landau, E. | title = Sur la Série des Invers de Nombres de Fibonacci | journal = Bull. Soc. Math. France | volume = 27 | year = 1899 | pages = 298–300}}
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Infinite sums over [[multiplicative inverse|reciprocal]] Fibonacci numbers can sometimes be evaluated in terms of [[theta function]]s. For example, the sum of every odd-indexed reciprocal Fibonacci number can be written as
<math display=block>\sum_{k=1}^\infty \frac{1}{F_{2 k-1}} = \frac{\sqrt{5}}{4} \; \vartheta_2\!\left(0, \frac{3-\sqrt 5}{2}\right)^2 ,</math>

and the sum of squared reciprocal Fibonacci numbers as
<math display=block>\sum_{k=1}^\infty \frac{1}{{F_k}^2} = \frac{5}{24} \!\left(\vartheta_2\!\left(0, \frac{3-\sqrt 5}{2}\right)^4 - \vartheta_4\!\left(0, \frac{3-\sqrt 5}{2}\right)^4 + 1 \right).</math>

If we add 1 to each Fibonacci number in the first sum, there is also the closed form
<math display=block>\sum_{k=1}^\infty \frac{1}{1+F_{2 k-1}} = \frac{\sqrt{5}}{2},</math>

and there is a ''nested'' sum of squared Fibonacci numbers giving the reciprocal of the [[golden ratio]],
<math display=block>\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2} .</math>

The sum of all even-indexed reciprocal Fibonacci numbers is<ref>[[Edmund Landau|Landau]] (1899)<!-- most probably: {{Citation | author = Landau, E. | title = Sur la Série des Invers de Nombres de Fibonacci | journal = Bull. Soc. Math. France | volume = 27 | year = 1899 | pages = 298–300}}
--> quoted according [[#Borwein|Borwein]], Page 95, Exercise 3b.</ref>
<math display=block>\sum_{k=1}^{\infty} \frac{1}{F_{2 k}} = \sqrt{5} \left(L(\psi^2) - L(\psi^4)\right) </math>
with the [[Lambert series]] <math>\textstyle L(q) := \sum_{k=1}^{\infty} \frac{q^k}{1-q^k} ,</math> since <math>\textstyle \frac{1}{F_{2 k}} = \sqrt{5} \left(\frac{\psi^{2 k}}{1-\psi^{2 k}} - \frac{\psi^{4 k}}{1-\psi^{4 k}} \right)\!.</math>

So the [[reciprocal Fibonacci constant]] is<ref>{{Cite OEIS|1=A079586|2=Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the {{mvar|k}}-th Fibonacci number|mode=cs2}}</ref>
<math display=block>\sum_{k=1}^{\infty} \frac{1}{F_k} = \sum_{k=1}^\infty \frac{1}{F_{2 k-1}} + \sum_{k=1}^{\infty} \frac {1}{F_{2 k}} = 3.359885666243 \dots</math>

Moreover, this number has been proved [[irrational number|irrational]] by [[Richard André-Jeannin]].<ref>{{citation
 | last = André-Jeannin
 | first = Richard
 | title = Irrationalité de la somme des inverses de certaines suites récurrentes
 | journal = [[Comptes Rendus de l'Académie des Sciences, Série I]]
 | volume = 308
 | year = 1989
 | issue = 19
 | pages = 539–41
 |mr=0999451}}</ref>

'''Millin's series''' gives the identity<ref>{{citation|title=Mathematical Gems III|volume=9|series=Dolciani Mathematical Expositions|first=Ross|last=Honsberger|publisher=American Mathematical Society|year=1985|isbn=9781470457181|contribution=Millin's series|pages=135–136|contribution-url=https://books.google.com/books?id=vl_0DwAAQBAJ&pg=PA135}}</ref>
<math display=block>\sum_{k=0}^{\infty} \frac{1}{F_{2^k}} = \frac{7 - \sqrt{5}}{2},</math>
which follows from the closed form for its partial sums as {{mvar|N}} tends to infinity:
<math display=block>\sum_{k=0}^N \frac{1}{F_{2^k}} = 3 - \frac{F_{2^N-1}}{F_{2^N}}.</math>