On the reciprocal sum of the fourth power of Fibonacci numbers

Abstract

Abstract Let f n {f}_{n} be the n n th Fibonacci number with f 1 = f 2 = 1 {f}_{1}={f}_{2}=1 . Recently, the exact values of ∑ k = n ∞ 1 f k s − 1 ⌊{\left({\sum }_{k=n}^{\infty }\frac{1}{{f}_{k}^{s}}\right)}^{-1}⌋ have been obtained only for s = 1 , 2 s=1,2 , where ⌊ x ⌋ \lfloor x\rfloor is the floor function. It has been an open problem for s ≥ 3 s\ge 3 . In this article, we consider the case of s = 4 s=4 and show that ∑ k = n ∞ 1 f k 4 − 1 = f n 4 − f n − 1 4 + 2 ( − 1 ) n 5 f 2 n − 1 − n + 2 5 , ⌊{\left(\mathop{\sum }\limits_{k=n}^{\infty }\frac{1}{{f}_{k}^{4}}\right)}^{-1}⌋={f}_{n}^{4}-{f}_{n-1}^{4}+\frac{2{\left(-1)}^{n}}{5}{f}_{2n-1}-\left\{\frac{n+2}{5}\right\}, where { x } ≔ x − ⌊ x ⌋ \left\{x\right\}:= x-\lfloor x\rfloor .