| Summary: | Let fn{f}_{n} be the nnth Fibonacci number with f1=f2=1{f}_{1}={f}_{2}=1. Recently, the exact values of ∑k=n∞1fks−1⌊{\left({\sum }_{k=n}^{\infty }\frac{1}{{f}_{k}^{s}}\right)}^{-1}⌋ have been obtained only for s=1,2s=1,2, where ⌊x⌋\lfloor x\rfloor is the floor function. It has been an open problem for s≥3s\ge 3. In this article, we consider the case of s=4s=4 and show that ∑k=n∞1fk4−1=fn4−fn−14+2(−1)n5f2n−1−n+25,⌊{\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 .
|
|---|