Editing Theta function (section)

===Sums with theta function in the result===

The infinite sum<ref>Landau (1899)<!-- Vermutl: {{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}}
--> zitiert nach [[Fibonacci-Folge#Borwein|Borwein]], Page 94, Exercise 3.</ref><ref>{{cite web|access-date=2021-07-18|title=Number-theoretical, combinatorial and integer functions – mpmath 1.1.0 documentation|url=https://mpmath.org/doc/1.1.0/functions/numtheory.html}}<!-- auto-translated by Module:CS1 translator --></ref> of the reciprocals of [[Fibonacci sequence|Fibonacci numbers]] with odd indices has the identity:

:<math>\sum_{n=1}^\infty \frac{1}{F_{2n-1}} = \frac{\sqrt{5}}{2}\,\sum_{n=1}^\infty \frac{2(\Phi^{-2})^{n - 1/2}}{1 + (\Phi^{-2})^{2n - 1}} = \frac{\sqrt{5}}{4} \sum_{a=-\infty}^\infty \frac{2(\Phi^{-2})^{a - 1/2}}{1 + (\Phi^{-2})^{2a - 1}} =</math>
:<math>= \frac{\sqrt{5}}{4}\,\theta_{2}(\Phi^{-2})^2 = \frac{\sqrt{5}}{8}\bigl[\theta_{3}(\Phi^{-1})^2 - \theta_{4}(\Phi^{-1})^2\bigr]</math>

By not using the theta function expression, following identity between two sums can be formulated:

:<math>\sum_{n=1}^\infty \frac{1}{F_{2n-1}} = \frac{\sqrt{5}}{4}\,\biggl[ \sum_{n=1}^\infty 2 \,\Phi^{- (2n - 1)^2 /2} \biggr]^2 </math>

:<math>\sum_{n=1}^\infty \frac{1}{F_{2n-1}} = 1.82451515740692456814215840626732817332\ldots </math>

Also in this case <math> \Phi = \tfrac{1}{2}(\sqrt{5} + 1)</math> is [[Golden ratio]] number again.

Infinite sum of the reciprocals of the Fibonacci number squares:
:<math>\sum_{n=1}^\infty \frac{1}{F_{n}^2} = \frac{5}{24}\bigl[2\,\theta_{2}(\Phi^{-2})^4 - \theta_{3}(\Phi^{-2})^4 + 1\bigr] = \frac{5}{24}\bigl[\theta_{3}(\Phi^{-2})^4 - 2\,\theta_{4}(\Phi^{-2})^4 + 1\bigr]</math>

Infinite sum of the reciprocals of the [[Pell sequence|Pell numbers]] with odd indices:
:<math>\sum_{n=1}^\infty \frac{1}{P_{2n-1}} = \frac{1}{\sqrt{2}}\,\theta_{2}\bigl[(\sqrt{2}-1)^2\bigr]^2 = \frac{1}{2\sqrt{2}}\bigl[\theta_{3}(\sqrt{2}-1)^2 - \theta_{4}(\sqrt{2}-1)^2\bigr]</math>