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Claim. $\sum_{k\ge0}1/F_{2^k}=(7-\sqrt5)/2$. The proof is the finite identity

$$\sum_{k=0}^{n}\frac{1}{F_{2^k}}=3-\frac{F_{2^n-1}}{F_{2^n}},$$

for $n\ge1$, verified by induction with Binet's formula, followed by letting $n\to\infty$, where $F_{2^n-1}/F_{2^n}$ tends to $(\sqrt5-1)/2$. The value is a quadratic irrational, so the instance $n_k=2^k$ of Problem 267 has answer yes. The source is I. J. Good, A reciprocal series of Fibonacci numbers, Fibonacci Quart. 12 (1974), no. 4, 346, on the card good_1974_reciprocal_series_fibonacci_numbers.

Covers. The instance $n_k=2^k$: the answer is yes. Not covered: every other index sequence. Hoggatt and Bicknell's 1976 evaluation (the Hoggatt–Bicknell page) generalizes the value to the indices $2^jk$ for every fixed $k$, and Badea's 1993 corollary (the Badea page) contains the instance in its condition.

Acceptance. Refereed: The Fibonacci Quarterly, volume 12, number 4 (December 1974). The site's commentary credits Good, with Bicknell and Hoggatt, with the irrationality of this sum but labels the problem OPEN, so that commentary is not listed as reviewed evidence. The corpus has not reproved the identity and awards no tier of its own.

Depends on. Nothing in this wiki; the claim rests on the cited paper.