linguistic:paraphrasing, stylistic:tone
Reciprocal sums

Infinite sums over reciprocal Fibonacci numbers can sometimes be evaluated in terms of theta functions. For example, the sum of every odd-indexed reciprocal Fibonacci number can be written as ∑ k = 1 ∞ 1 F 2 k − 1 = 5 4 ϑ 2 (0, 3 − 5 2) 2, {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{2k-1}}}={\frac {\sqrt {5}}{4}}\;\vartheta _{2}\!\left(0,{\frac {3-{\sqrt {5}}}{2}}\right)^{2},}

and the sum of squared reciprocal Fibonacci numbers as ∑ k = 1 ∞ 1 F k 2 = 5 24 (ϑ 2 (0, 3 − 5 2) 4 − ϑ 4 (0, 3 − 5 2) 4 + 1). {\displaystyle \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).}

If we add 1 to each Fibonacci number in the first sum, there is also the closed form ∑ k = 1 ∞ 1 1 + F 2 k − 1 = 5 2, {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{1+F_{2k-1}}}={\frac {\sqrt {5}}{2}},}

and there is a nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio, ∑ k = 1 ∞ (− 1) k + 1 ∑ j = 1 k F j 2 = 5 − 1 2. {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{\sum _{j=1}^{k}{F_{j}}^{2}}}={\frac {{\sqrt {5}}-1}{2}}.}

The sum of all even-indexed reciprocal Fibonacci numbers is ∑ k = 1 ∞ 1 F 2 k = 5 (L (ψ 2) − L (ψ 4)) {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{2k}}}={\sqrt {5}}\left(L(\psi ^{2})-L(\psi ^{4})\right)} with the Lambert series L (q):= ∑ k = 1 ∞ q k 1 − q k, {\displaystyle \textstyle L(q):=\sum _{k=1}^{\infty }{\frac {q^{k}}{1-q^{k}}},} since 1 F 2 k = 5 (ψ 2 k 1 − ψ 2 k − ψ 4 k 1 − ψ 4 k). {\displaystyle \textstyle {\frac {1}{F_{2k}}}={\sqrt {5}}\left({\frac {\psi ^{2k}}{1-\psi ^{2k}}}-{\frac {\psi ^{4k}}{1-\psi ^{4k}}}\right)\!.}

So the reciprocal Fibonacci constant is ∑ k = 1 ∞ 1 F k = ∑ k = 1 ∞ 1 F 2 k − 1 + ∑ k = 1 ∞ 1 F 2 k = 3.359885666243 … {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{k}}}=\sum _{k=1}^{\infty }{\frac {1}{F_{2k-1}}}+\sum _{k=1}^{\infty }{\frac {1}{F_{2k}}}=3.359885666243\dots }

Moreover, this number has been proved irrational by Richard André-Jeannin.

Millin's series gives the identity ∑ k = 0 ∞ 1 F 2 k = 7 − 5 2, {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{F_{2^{k}}}}={\frac {7-{\sqrt {5}}}{2}},} which follows from the closed form for its partial sums as N tends to infinity: ∑ k = 0 N 1 F 2 k = 3 − F 2 N − 1 F 2 N. {\displaystyle \sum _{k=0}^{N}{\frac {1}{F_{2^{k}}}}=3-{\frac {F_{2^{N}-1}}{F_{2^{N}}}}.}