By Lemma 1(i), we have that

$$ F_{mk}= \frac{\alpha ^{mk}}{\sqrt{5}} \biggl(1- \biggl( \frac{\beta }{\alpha } \biggr)^{mk} \biggr), $$

hence

$$ \frac{1}{F_{mk}^{2}}= \frac{5}{\alpha ^{2mk}(1-(\beta /\alpha )^{mk})^{2}}. $$

However, \(1/(1-x)^{2}=1+2x+3x^{2}+\cdots \) holds for \(|x|<1\). Thus, since \(|\beta /\alpha |<1\) and \(\alpha \beta =-1\), we have

$$\begin{aligned} \frac{1}{F_{mk}^{2}} = & \frac{5}{\alpha ^{2mk}} \biggl(1+2 \biggl( \frac{\beta }{\alpha } \biggr)^{mk}+3 \biggl( \frac{\beta }{\alpha } \biggr)^{2mk} + \cdots \biggr) \\ = & \frac{5}{\alpha ^{2mk}} + 5 \sum_{i=2}^{ \infty } i (-1)^{mk} \biggl( \frac{\beta }{\alpha } \biggr)^{imk}. \end{aligned}$$

By summing up from \(k=n\) to infinity and after a straightforward calculation, we arrive at

$$\begin{aligned} \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} = & \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}+5 \sum_{i=2}^{\infty } i\sum_{k=n}^{\infty } (-1)^{mk} \biggl( \frac{\beta }{\alpha } \biggr)^{imk} \\ = & \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)} + 5 \sum_{i=2}^{\infty } i (-1)^{mn} \biggl( \frac{\beta }{\alpha } \biggr)^{imn} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{im}} \\ = & \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}\bigl(1+f(m,n)\bigr), \end{aligned}$$

(7)

where \(f(m,n)\) denotes the following summatory

$$ f(m,n):= \biggl( \frac{\alpha ^{2m}-1}{\alpha ^{2m}} \biggr) \sum_{i=2}^{\infty } i \biggl( \frac{\beta }{\alpha } \biggr)^{mn(i-1)} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{im}} . $$

Note that, by the reverse triangle inequality (and \(\beta /\alpha =-1/\alpha ^{2}\), which follows from \(\beta =-1/\alpha \)), one has

$$ \bigl\vert 1-(-1)^{m}(\beta /\alpha )^{im} \bigr\vert \geq 1- \biggl\vert \frac{\beta }{\alpha } \biggr\vert ^{2m}>1- \frac{1}{\alpha ^{4}}> 0.85 . $$

Thus, we deduce the following upper bound for \(|f(m,n)|\):

$$\begin{aligned} \bigl\vert f(m,n) \bigr\vert \leq & \frac{1}{0.85} \frac{\alpha ^{2m}-1}{\alpha ^{2m}} \sum_{i=2}^{\infty } i \biggl( \frac{1}{\alpha ^{2}} \biggr)^{mn(i-1)} \\ \leq & \frac{1.18}{\alpha ^{2mn}} \sum_{i=2}^{ \infty } i \biggl( \frac{1}{\alpha ^{2mn}} \biggr)^{i-2} \\ = & \frac{1.18}{\alpha ^{2mn}} \Biggl(2 + \frac{3}{\alpha ^{2mn}} + \sum _{i=4}^{\infty } \frac{i}{\alpha ^{2mn(i-2)}} \Biggr) \\ = & \frac{1.18}{\alpha ^{2mn}} \Biggl(2 + \frac{3}{\alpha ^{2mn}} + \sum _{i=3}^{\infty } \frac{i+1}{\alpha ^{mn(2i-2)}} \Biggr) \\ = & \frac{1.18}{\alpha ^{2mn}} \Biggl(2 + \frac{3}{\alpha ^{2mn}} + \sum _{i=3}^{\infty } \biggl( \frac{i+1}{\alpha ^{imn}} \cdot \frac{1}{\alpha ^{(i-2)mn}} \biggr) \Biggr) \\ \leq & \frac{1.18}{\alpha ^{2mn}} \Biggl(2.065 + \frac{1}{60} \sum _{i=1}^{\infty } \frac{1}{\alpha ^{imn}} \Biggr) \\ \leq & \frac{1.18}{\alpha ^{2mn}} \biggl(2.065 + \frac{0.171}{60} \biggr) \\ < & \frac{2.45}{\alpha ^{2mn}}, \end{aligned}$$

where we used that

$$ \alpha ^{imn} > 1.6^{imn}\geq 6.55^{i} > 60(i+1) $$

for all \(i\geq 3\) (since \(mn\geq 4\)) together with

$$ \frac{1}{\alpha ^{mn}}+ \frac{1}{\alpha ^{2mn}}+ \frac{1}{\alpha ^{3mn}}+\cdots < \frac{1}{\alpha ^{4}}+ \frac{1}{\alpha ^{8}}+ \frac{1}{\alpha ^{12}}+\cdots = \frac{1}{\alpha ^{4}-1}< 0.171 , $$

because \(\alpha ^{mn}\geq \alpha ^{4}\). Thus \(f(m,n)\) tends to 0 as \(\min \{m,n\}\to \infty \). Furthermore, \(|f(m,n)|<0.06\) for all integers *m* and *n* with \(mn\geq 4\).

Turning back to (7), we have that

$$ \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}}= \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}\cdot \bigl(1+f(m,n)\bigr), $$

and hence

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5} \biggl( \frac{1}{1+f(m,n)} \biggr). $$

(8)

Since \(|f(m,n)|<1\), for \(mn\geq 4\), then

$$ \frac{1}{1+f(m,n)}=1-f(m,n)+\bigl(f(m,n)\bigr)^{2}-\cdots =1-f(m,n)+R_{m,n}, $$

where \(R_{m,n}:=(f(m,n))^{2}-(f(m,n))^{3}+(f(m,n))^{4}-\cdots \) . For our purposes, we need to find an upper bound for \(|R_{m,n}|\). For that, one has

$$\begin{aligned} \vert R_{m,n} \vert \leq & \bigl\vert \bigl(f(m,n)\bigr)^{2}\bigl(1-f(m,n)+\bigl(f(m,n)\bigr)^{2}- \cdots \bigr) \bigr\vert \\ = & \biggl\vert \frac{(f(m,n))^{2}}{1+f(m,n)} \biggr\vert \\ \leq & \frac{(f(m,n))^{2}}{1- \vert f(m,n) \vert } \\ \leq & 1.07 \bigl(f(m,n)\bigr)^{2} < \frac{6.4}{\alpha ^{4mn}}, \end{aligned}$$

(9)

where we used that

$$ \bigl\vert 1+f(m,n) \bigr\vert \geq 1- \bigl\vert f(m,n) \bigr\vert >1- \frac{2.37}{\alpha ^{2mn}}>1-0.06=0.94. $$

Turning back to (8), we have

$$\begin{aligned} \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} = & \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}\bigl(1-f(m,n)+R_{m,n} \bigr) \\ = & \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n) \\ &{} + \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5} R_{m,n} \\ = & \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n) \\ & + T_{m,n} , \end{aligned}$$

where

$$ T_{m,n}:= \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5} \cdot R_{m,n} $$

satisfies (by (9))

$$ \vert T_{m,n} \vert = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1) \vert R_{m,n} \vert }{5}< \frac{\alpha ^{2mn}}{5}\cdot \frac{6.4}{\alpha ^{4mn}}< \frac{1.3}{\alpha ^{2mn}}. $$

(10)

Hence

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1}= \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n)+T_{m,n}. $$

(11)

Now, let us work with the second term of the right-hand side of (11). By the definition of \(f(m,n)\), we can write

$$\begin{aligned}& \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n) \\& \quad =\frac{2\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \biggl( \frac{\beta }{\alpha } \biggr)^{mn} \frac{1}{1-(-1)^{m} (\beta /\alpha )^{2m}} + D_{m,n}, \end{aligned}$$

where

$$ D_{m,n}:= \frac{\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \sum_{i=3}^{\infty } i \biggl( \frac{\beta }{\alpha } \biggr)^{mn(i-1)} \frac{1}{1-(-1)^{m} (\beta /\alpha )^{im}} . $$

Since \(\beta /\alpha =-1/\alpha ^{2}\), we deduce that

$$\begin{aligned}& \frac{2\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \biggl( \frac{\beta }{\alpha } \biggr)^{mn} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{2m}} \\& \quad = \frac{2\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \biggl( \frac{-1}{\alpha ^{2}} \biggr)^{mn} \frac{1}{1-(-1/\alpha ^{4})^{m}} \\& \quad = \frac{2(-1)^{mn}}{5} \cdot \frac{\alpha ^{4m} - 2\alpha ^{2m} + 1}{\alpha ^{4m}} \Biggl(1+ \sum _{i=1}^{ \infty } \biggl( \frac{-1}{\alpha ^{4}} \biggr)^{mi} \Biggr) \\& \quad = \frac{2(-1)^{mn}}{5}(1-S_{m}) (1+E_{m}) \\& \quad = \frac{2(-1)^{mn}}{5}+(-1)^{mn}G_{m}, \end{aligned}$$

where

$$ G_{m}:= \frac{2}{5}(E_{m}-S_{m}-E_{m}S_{m}) $$

(12)

and \(E_{m}\) and \(S_{m}\) are defined as

$$ S_{m}:= \frac{2}{\alpha ^{2m}}- \frac{1}{\alpha ^{4m}}\quad \text{and}\quad E_{m}:=\sum_{i=1}^{\infty } \biggl( \frac{-1}{\alpha ^{4}} \biggr)^{mi} = \frac{(-1)^{m}}{\alpha ^{4m}-(-1)^{m}}. $$

(13)

Now, we use that \(S_{m}<2/\alpha ^{2m}\) and

$$ \vert E_{m} \vert \leq \sum_{i=1}^{\infty } \biggl\vert \frac{-1}{\alpha ^{4}} \biggr\vert ^{mi} = \frac{1}{\alpha ^{4m}-1} = \frac{1}{\alpha ^{4m}} \frac{\alpha ^{4m}}{\alpha ^{4m}-1} \leq \frac{1}{\alpha ^{4m}} \frac{\alpha ^{4}}{\alpha ^{4}-1} < \frac{1.2}{\alpha ^{4m}} $$

(since \(x\mapsto x/(x-1)\) is a decreasing function for \(x>1\), and so the maximum of \(\alpha ^{4m}/(\alpha ^{4m}-1)\) is attained at \(m=1\)) to infer that

$$ \vert G_{m} \vert \leq \frac{2}{5} \bigl( \vert E_{m} \vert + \vert S_{m} \vert + \vert E_{m}S_{m} \vert \bigr)\leq \frac{2}{5} \biggl( \frac{1.2}{\alpha ^{4m}}+ \frac{2}{\alpha ^{2m}}+ \frac{2.4}{\alpha ^{6m}} \biggr)\leq \frac{1.2}{\alpha ^{2m}}, $$

(14)

which proves (6). Observe that, in particular, \(G_{m}\) tends to 0 as \(m\to \infty \).

For the remaining terms, i.e., \(D_{m,n}\), first we can realize that

$$ \frac{\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} < \frac{\alpha ^{2mn}}{5} , $$

(15)

since \((\alpha ^{2m}-1)^{2}<\alpha ^{4m}\). Therefore, by using again \(\beta /\alpha =-1/\alpha ^{2}\), we get

$$\begin{aligned} \Biggl\vert \sum_{i=3}^{\infty } i \frac{ (\beta /\alpha ) ^{mn(i-1)}}{1-(-1)^{m} (\beta /\alpha )^{im}} \Biggr\vert = & \Biggl\vert \sum _{i=3}^{\infty } i \biggl( \frac{-1}{\alpha ^{2}} \biggr)^{mn(i-1)} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{im}} \Biggr\vert \\ < & 1.32 \cdot \sum_{i=3}^{\infty } i \frac{1}{\alpha ^{2mn(i-1)}} \\ < & 1.32 \cdot \sum_{i=3}^{\infty } \frac{1}{\alpha ^{2mn(i-1)-2i}}, \end{aligned}$$

where we used that \(i<2^{i}<\alpha ^{2i}\) holds for every integer \(i\geq 3\). Furthermore, since \(mn\geq 4\), we have for \(i\geq 2\)

$$ a^{2mn(i-1)-2i} \geq \bigl(\alpha ^{2mn (\frac{i-1}{i} )-2} \bigr)^{i} \geq \bigl(\alpha ^{ \frac{2mn}{2}-2} \bigr)^{i} \geq \bigl(\alpha ^{2} \bigr)^{i}> 2.6^{i} . $$

Thus,

$$\begin{aligned} &\Biggl\vert \sum_{i=3}^{\infty } i \frac{ (\beta /\alpha ) ^{mn(i-1)}}{1-(-1)^{m} (\beta /\alpha )^{im}} \Biggr\vert \\ &\quad \leq 1.32\cdot \sum_{i=3}^{\infty } \frac{1}{\alpha ^{2mn(i-1)-2i}} = \frac{1.32}{\alpha ^{4mn-4}} \sum_{i=3}^{\infty } \frac{1}{\alpha ^{2mn(i-3)-2(i-2)}} \\ &\quad = \frac{1.32}{\alpha ^{4mn-4}} \sum_{i=1}^{\infty } \frac{1}{\alpha ^{2mn(i-1)-2i}} = \frac{1.32}{\alpha ^{4mn-4}} \Biggl(\frac{1}{\alpha ^{-2}} + \sum _{i=2}^{ \infty } \frac{1}{\alpha ^{2mn(i-1)-2i}} \Biggr) \\ &\quad < \frac{1.32}{\alpha ^{4mn-4}} \Biggl( \frac{1}{\alpha ^{-2}} + \sum _{i=2}^{\infty } \frac{1}{2.6^{i}} \Biggr) \\ &\quad < \frac{3.78}{\alpha ^{4mn-4}}. \end{aligned}$$

(16)

By combining (15) and (16), we infer that

$$ \vert D_{m,n} \vert < \frac{0.76}{\alpha ^{2mn-4}}. $$

(17)

In particular,

$$ \lim_{\min \{m,n\}\to \infty }D_{m,n}=0. $$

Summarizing, we have

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1}= \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{2(-1)^{mn}}{5}-(-1)^{mn}G_{m}+D_{m,n}+T_{m,n}. $$

(18)

On the other hand, we can apply Binet’s formula again to obtain

$$\begin{aligned} F_{mn}^{2}- F_{m(n-1)}^{2}& = \biggl( \frac{\alpha ^{mn}-\beta ^{mn}}{\sqrt{5}} \biggr)^{2} + \biggl( \frac{\alpha ^{m(n-1)}-\beta ^{m(n-1)}}{\sqrt{5}} \biggr)^{2} \\ & = \frac{\alpha ^{2mn}-2(-1)^{mn}+\beta ^{2mn}-\alpha ^{2m(n-1)}+2(-1)^{m(n-1)}-\beta ^{2m(n-1)}}{5} \\ & = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{2(-1)^{mn}}{5}\bigl(1+(-1)^{m}\bigr) + \biggl( \frac{\beta ^{2mn}-\beta ^{2m(n-1)}}{5} \biggr), \\ & = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{2(-1)^{mn}}{5}- \frac{2(-1)^{m(n-1)}}{5} - \biggl( \frac{\alpha ^{2m}-1}{5\alpha ^{2mn}} \biggr), \end{aligned}$$

where we used that \(\alpha \beta =-1\). Now, we can combine the previous relation with the formula in (18) to write

$$ A_{m,n}= \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1}- \bigl(F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m} \bigr), $$

(19)

where

$$ C_{m}:= \frac{2}{5}+(-1)^{m}G_{m}\quad \text{and}\quad A_{m,n}:= \frac{1-\alpha ^{2m}}{5\alpha ^{2mn}} + D_{m,n}+T_{m,n} . $$

Now, we use the formula for \(A_{m,n}\), estimates (17) and (10) to obtain

$$ \vert A_{m,n} \vert < \frac{0.2}{\alpha ^{2m(n-1)}}+ \frac{5.3}{\alpha ^{2mn}}+ \frac{1.3}{\alpha ^{2mn}}= \frac{1}{\alpha ^{2m(n-1)}} \biggl(0.2+ \frac{5.3}{\alpha ^{2m}}+ \frac{1.3}{\alpha ^{2m}} \biggr)< \frac{9.83}{\alpha ^{2m(n-1)}}, $$

which implies in (3) (we used that \(0.76/\alpha ^{2mn-4}=0.76\alpha ^{4}/\alpha ^{2mn}\)). Moreover, in particular, \(A_{m,n}\) tends to 0 as \(n\to \infty \), then (4) holds, i.e.,

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \sim F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m}. $$

Moreover, \(C_{m}\) is positive, because by (14) one has

$$ C_{m}:= \frac{2}{5}+(-1)^{m}G_{m}\geq \frac{2}{5}- \vert G_{m} \vert > \frac{2}{5}- \frac{1.2}{\alpha ^{2m}}>0, $$

where we used that \(1.2/\alpha ^{2m}\leq 1.2/\alpha ^{4}<0.18\) for all \(m\geq 2\). Additionally, we combine (12) and (13) to obtain

$$\begin{aligned} C_{m} =& \frac{2}{5} \biggl( 1 + \frac{1}{\alpha ^{4m}-(-1)^{m}}- \frac{2(-1)^{m}}{\alpha ^{2m}}+ \frac{(-1)^{m}}{\alpha ^{4m}} \\ &{}- \frac{2}{\alpha ^{2m}(\alpha ^{4m}-(-1)^{m})}+ \frac{1}{\alpha ^{4m}(\alpha ^{4m}-(-1)^{m})} \biggr). \end{aligned}$$

(20)

To obtain the formula in (5), we shall rationalize every fraction in (20). First, we observe that \((\alpha \beta )^{im}=1\) for \(i\in \{2,4\}\) and

$$ \bigl(\alpha ^{4m}-(-1)^{m}\bigr) \bigl(\beta ^{4m}-(-1)^{m}\bigr)=(\alpha \beta )^{4m}-(-1)^{m} \bigl( \alpha ^{4m}+\beta ^{4m}\bigr)+1=2-(-1)^{m}L_{4m}. $$

Now, let us work with the expression

$$ \frac{1}{\alpha ^{4m}-(-1)^{m}}- \frac{2(-1)^{m}}{\alpha ^{2m}}+ \frac{(-1)^{m}}{\alpha ^{4m}}- \frac{2}{\alpha ^{2m}(\alpha ^{4m}-(-1)^{m})}+ \frac{1}{\alpha ^{4m}(\alpha ^{4m}-(-1)^{m})}. $$

(21)

By rationalizing, one has

$$\begin{aligned}& \frac{1}{\alpha ^{4m}-(-1)^{m}}= \frac{\beta ^{4m}-(-1)^{m}}{2-(-1)^{m}L_{m}} \quad \text{and}\quad \frac{2(-1)^{m}}{\alpha ^{2m}}=2(-1)^{m} \beta ^{2m}, \\& \frac{(-1)^{m}}{\alpha ^{4m}}=(-1)^{m}\beta ^{4m}\quad \text{and}\quad \frac{2}{\alpha ^{2m}(\alpha ^{4m}-(-1)^{m})}= \frac{2\beta ^{2m}(\alpha ^{4m}-(-1)^{m})}{2-(-1)^{m}L_{4m}}, \end{aligned}$$

and finally

$$ \frac{1}{\alpha ^{4m}(\alpha ^{4m}-(-1)^{m})}= \frac{\beta ^{4m}(\beta ^{4m}-(-1)^{m})}{2-(-1)^{m}L_{4m}}. $$

By putting all this information together and after a straightforward computation, we deduce that

$$ C_{m}= \frac{2}{5} \biggl(1 + \frac{B_{m}}{(2-(-1)^{m}L_{4m})} \biggr), $$

where

$$\begin{aligned} B_{m} := & \beta ^{4m}-(-1)^{m}-4(-1)^{m} \beta ^{2m}+ 2\beta ^{2m}L_{4m}+2(-1)^{m} \beta ^{4m} \\ & {}-\beta ^{4m}L_{4m}-2\beta ^{6m}+2(-1)^{m} \beta ^{2m}+\beta ^{8m}-(-1)^{m} \beta ^{4m}. \end{aligned}$$

We then use \(\beta ^{s}=\beta F_{s}+F_{s-1}\) (Lemma 3) to write \(B_{m}\) in the form \(r_{m}+\beta s_{m}\) (where \(r_{m}\) and \(s_{m}\) belong to \(\mathbb{Q}\) and they are called the rational and irrational parts of \(B_{m}\), respectively). After some manipulations (by using \(F_{2\ell m}=F_{\ell m}L_{\ell m}\), where *ℓ* is a positive integer, and identity \(F_{4m-1}-1 = L_{2m-1}F_{2m}\), see [7]), we arrive at

$$ r_{m} := \bigl(1 - (-1)^{m}\bigr) L_{2 m} + 5 \bigl(1 + (-1)^{m}\bigr) F_{m - 1} F_{m} F_{2 m} $$

(22)

and

$$ s_{m}:=\bigl(1+(-1)^{m}\bigr)F_{2m}(L_{2m}-2) $$

(23)

as desired. The proof is then complete.