Editing Theta function (section)

==Some series identities==

===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>

===Sums with theta function in the summand===
The next two series identities were proved by [[István Mező]]:<ref name="Mezo2">{{Citation
  | last = Mező
  | first = István
  | title = Duplication formulae involving Jacobi theta functions and Gosper's {{mvar|q}}-trigonometric functions
  | journal = [[Proceedings of the American Mathematical Society]]
  | pages = 2401–2410
  | volume = 141
  | issue = 7
  | year = 2013
  | doi=10.1090/s0002-9939-2013-11576-5 | doi-access = free}}</ref>

:<math>\begin{align}
\theta_4^2(q)&=iq^{\frac14}\sum_{k=-\infty}^\infty q^{2k^2-k}\theta_1\left(\frac{2k-1}{2i}\ln q,q\right),\\[6pt]
\theta_4^2(q)&=\sum_{k=-\infty}^\infty q^{2k^2}\theta_4\left(\frac{k\ln q}{i},q\right).
\end{align}</math>

These relations hold for all {{math|0 < ''q'' < 1}}. Specializing the values of {{mvar|q}}, we have the next parameter free sums

:<math>\sqrt{\frac{\pi\sqrt{e^\pi}}{2}}\cdot\frac{1}{\Gamma^2\left(\frac34\right)} =i\sum_{k=-\infty}^\infty e^{\pi\left(k-2k^2\right)} \theta_1 \left(\frac{i\pi}{2}(2k-1),e^{-\pi}\right)</math>

:<math>\sqrt{\frac{\pi}{2}}\cdot\frac{1}{\Gamma^2\left(\frac34\right)} =\sum_{k=-\infty}^\infty\frac{\theta_4\left(ik\pi,e^{-\pi}\right)}{e^{2\pi k^2}}</math>