Franklin’s Blog

Reciprocal-Fibonacci Series

2018 Jan 30

Evaluate the following infinite series, where \(F_n\) is the nth fibonacci number with \(F_1=F_2=1\):
\[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{F_k F_{k+5}}=\space\space\space ?\]
\[\sum_{k=1}^\infty \frac{1}{F_k F_{k+6}}=\space\space\space ?\]
\[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{F_{5k-1}F_{5k+4}}=\space\space\space ?\]
\[\sum_{k=1}^\infty \frac{1}{F_{10k}F_{10k+10}}=\space\space\space ?\]

To evaluate these series, consider the functions \(\Phi_n(x)\) defined as
\[\Phi_n(x):=\sum_{k=1}^\infty \frac{x^{k+1}}{F_k F_{k+n}}\]
where \(F_k\) are the Fibonacci numbers, with \(F_1=F_2=1\). In this post, I shall endeavour to evaluate a few particular value of \(\Phi_n(x)\), establish a few of its functional equations to make computation easier, and use the functional equations to evaluate related sums.

First, recall the following well-known identities:
\[F_{k+n}=F_k F_{n-1}+F_{k+1}F_n\]
\[F_k^2-F_{k+1}F_{k-1}=(-1)^{k+1}\]
which I will use frequently throughout the post.

I shall begin by computing \(\Phi_1(-1)\) and \(\Phi_2(1)\), which are both straightforward telescoping sums:
\[\begin{align} \Phi_1(-1) &= \sum_{k=1}^\infty \frac{(-1)^{k+1}}{F_k F_{k+1}}\\ &= 1+\sum_{k=2}^\infty \frac{F_k^2-F_{k+1}F_{k-1}}{F_k F_{k+1}}\\ &= 1-\sum_{k=2}^\infty \frac{F_{k-1}}{F_{k}}-\frac{F_k}{F_{k+1}}\\ &= 1-1+\lim_{k\to\infty} \frac{F_k}{F_{k+1}}\\ &= \phi^{-1}\\ \end{align}\]
Where \(\phi\) is the golden ratio \(\frac{1+\sqrt 5}{2}\). Next, we have
\[\begin{align} \Phi_2(1) &= \sum_{k=1}^\infty \frac{1}{F_k F_{k+2}}\\ &= \sum_{k=1}^\infty \frac{1}{F_k F_{k+1}}-\frac{1}{F_{k+1}F_{k+2}}\\ &= 1+\lim_{k\to\infty} \frac{1}{F_{k+1}F_{k+2}}\\ &= 1\\ \end{align}\]

Now I will establish a recurrence relation that allows one to compute additional values of \(\Phi_n(x)\) using the two previously computed as initial conditions.

Theorem 1:
\[\Phi_n(x)=\frac{\Phi_1(x)}{F_n}-\frac{F_{n-1}}{F_n}\frac{\Phi_{n-1}(x)}{x}+\frac{F_{n-1}x}{F_n^2}\]

Proof:

\[\begin{align} \Phi_n(x) &=\sum_{k=1}^\infty \frac{x^{k+1}}{F_k F_{k+n}}\\ &=\sum_{k=1}^\infty \frac{x^{k+1}}{F_k (F_k F_{n-1}+F_{k+1}F_n)}\\ &=\sum_{k=1}^\infty \frac{x^{k+1}}{F_{k+1}F_n}\bigg(\frac{1}{F_k}-\frac{F_{n-1}}{F_{k+n}}\bigg)\\ &=\frac{\Phi_1(x)}{F_n}-\frac{F_{n-1}}{F_n}\sum_{k=1}^\infty \frac{x^{k+1}}{F_{k+1}F_{k+n}}\\ &=\frac{\Phi_1(x)}{F_n}+\frac{F_{n-1}}{F_n}\bigg(\frac{x}{F_n}-\sum_{k=1}^\infty \frac{x^{k}}{F_{k}F_{k+n-1}}\bigg)\\ &=\frac{\Phi_1(x)}{F_n}+\frac{F_{n-1}}{F_n}\bigg(\frac{x}{F_n}-\frac{\Phi_{n-1}(x)}{x}\bigg)\\ &= \frac{\Phi_1(x)}{F_n}-\frac{F_{n-1}}{F_n}\frac{\Phi_{n-1}(x)}{x}+\frac{F_{n-1}x}{F_n^2}\space\space\space\space\space\space\blacksquare\\ \end{align}\]

Through this formula, one may recursively calculate many values of \(\Phi_n(1)\) and \(\Phi_{n}(-1)\). For example, using \(\Phi_1(-1)=\phi^{-1}\) as previously calculated,
\[\begin{align} \Phi_2(-1) &=\frac{\Phi_1(-1)}{F_2}-\frac{F_1}{F_2}\frac{\Phi_1(-1)}{-1}+\frac{F_1 (-1)}{F_2^2}\\ &=2\Phi_1(-1)-1\\ &=2\phi^{-1}-1\\ \end{align}\]

and

\[\begin{align} \Phi_3(-1) &=\frac{\Phi_1(-1)}{F_3}-\frac{F_2}{F_3}\frac{\Phi_2(-1)}{-1}+\frac{F_2 (-1)}{F_3^2}\\ &=\frac{\phi^{-1}}{2}+\frac{2\phi^{-1}-1}{2}-\frac{1}{4}\\ &=\frac{3\phi^{-1}}{2}-\frac{3}{4}\\ \end{align}\]

In fact, one may solve the recurrence relation to obtain the following:

Theorem 2:
\[\Phi_n(x)=\frac{\Phi_1(x)}{F_n}\sum_{k=0}^{n-1} (-x)^{-k}-\frac{1}{F_n}\sum_{k=0}^{n-1}\frac{(-x)^{1-k}F_{n-k-1}}{F_{n-k}},\space\space\space n\ge 1\]

Proof: The base case of this formula - with \(n=1\) - holds trivially, so I will now begin the induction step.

Suppose that
\[\Phi_m(x)=\frac{\Phi_1(x)}{F_m}\sum_{k=0}^{m-1} (-x)^{-k}-\frac{1}{F_m}\sum_{k=0}^{m-1}\frac{(-x)^{1-k}F_{m-k-1}}{F_{m-k}}\]
is true for some \(m\). Then it follows that
\[-\frac{F_{m}}{F_{m+1} x}\Phi_m(x)=-\frac{\Phi_1(x)}{F_{m+1}x}\sum_{k=0}^{m-1} (-x)^{-k}+\frac{1}{F_{m+1} x}\sum_{k=0}^{m-1}\frac{(-x)^{1-k}F_{m-k-1}}{F_{m-k}}\]
\[-\frac{F_{m}}{F_{m+1} x}\Phi_m(x)=\frac{\Phi_1(x)}{F_{m+1}}\sum_{k=1}^{m} (-x)^{-k}-\frac{1}{F_{m+1}}\sum_{k=0}^{m-1}\frac{(-x)^{-k}F_{m-k-1}}{F_{m-k}}\]
\[\frac{\Phi_1(x)}{F_{m+1}}-\frac{F_{m}}{F_{m+1} x}\Phi_m(x)=\frac{\Phi_1(x)}{F_{m+1}}\sum_{k=0}^{m} (-x)^{-k}-\frac{1}{F_{m+1}}\sum_{k=0}^{m-1}\frac{(-x)^{-k}F_{m-k-1}}{F_{m-k}}\]
\[\frac{\Phi_1(x)}{F_{m+1}}-\frac{F_{m}}{F_{m+1} x}\Phi_m(x)+\frac{F_m x}{F_{m+1}^2}=\frac{\Phi_1(x)}{F_{m+1}}\sum_{k=0}^{m} (-x)^{-k}-\frac{1}{F_{m+1}}\sum_{k=0}^{m-1}\frac{(-x)^{-k}F_{m-k-1}}{F_{m-k}}+\frac{F_m x}{F_{m+1}^2}\]
\[\frac{\Phi_1(x)}{F_{m+1}}-\frac{F_{m}}{F_{m+1} x}\Phi_m(x)+\frac{F_m x}{F_{m+1}^2}=\frac{\Phi_1(x)}{F_{m+1}}\sum_{k=0}^{m} (-x)^{-k}-\frac{1}{F_{m+1}}\sum_{k=1}^{m}\frac{(-x)^{1-k}F_{m-k}}{F_{m-k+1}}+\frac{F_m x}{F_{m+1}^2}\]
\[\frac{\Phi_1(x)}{F_{m+1}}-\frac{F_{m}}{F_{m+1} x}\Phi_m(x)+\frac{F_m x}{F_{m+1}^2}=\frac{\Phi_1(x)}{F_{m+1}}\sum_{k=0}^{m} (-x)^{-k}-\frac{1}{F_{m+1}}\sum_{k=0}^{m}\frac{(-x)^{1-k}F_{m-k}}{F_{m-k+1}}\]
By Theorem 1, the LHS is equal to \(\Phi_{m+1}(x)\), and so
\[\Phi_{m+1}(x)=\frac{\Phi_1(x)}{F_{m+1}}\sum_{k=0}^{m} (-x)^{-k}-\frac{1}{F_{m+1}}\sum_{k=0}^{m}\frac{(-x)^{1-k}F_{m-k}}{F_{m-k+1}}\]
Which shows that the formula holds for \(m+1\) as well. Thus, by induction, it holds for all \(n\ge 1\). \(\blacksquare\)

Corollary 1:
\[\Phi_n(-1)=\frac{n\phi^{-1}}{F_n}-\frac{1}{F_n}\sum_{k=0}^{n-1} \frac{F_k}{F_{k+1}},\space\space\space n\ge 1\]

Proof: This follows directly from Theorem 2, since
\[\sum_{k=0}^{n-1} 1^{-k}=n\]
and since \(\Phi_1(-1)=\phi^{-1}\). \(\blacksquare\)

Corollary 2:
\[\Phi_{2n}(1)=\frac{1}{F_{2n}}\sum_{k=0}^{2n-1}\frac{(-1)^{k}F_{2n-k-1}}{F_{2n-k}},\space\space\space n\ge 1\]

Proof: This follows directly from Theorem 2, since
\[\sum_{k=0}^{n-1} (-1)^{-k}=\begin{cases} 0 & \text{if n is even}\\ 1 & \text{if n is odd} \end{cases}\]

\(\blacksquare\)

Corollary 3:
\[\Phi_{2n+1}(1)=\frac{\Phi_1(1)}{F_{2n+1}}+\frac{1}{F_{2n+1}}\sum_{k=1}^{n}\frac{1}{F_{2k}F_{2k+1}}, \space\space\space n\in\mathbb N, \space n\ge 1\]

Proof: This follows directly from Theorem 2, since
\[\sum_{k=0}^{n-1} (-1)^{-k}=\begin{cases} 0 & \text{if n is even}\\ 1 & \text{if n is odd} \end{cases}\]

\(\blacksquare\)

Having derived these formulas, the answers to the first two proposed problems come easily:
\[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{F_k F_{k+5}}=\phi^{-1}-\frac{83}{150}\]
\[\sum_{k=1}^\infty \frac{1}{F_k F_{k+6}}=\frac{143}{960}\]

I will now demonstrate how to use \(\Phi_n(x)\) to evaluate a class of rather intimidating sums - sums in the form
\[\sum_{k=1}^\infty \frac{(-1)^{nk}}{F_{nk-m}F_{nk+n-m}},\space\space\text{with}\space\space\space 0\le m\le n-1\]
To evaluate these sums, I will make use of the properties of the complex roots of unity:
\[\omega_n := \cos \frac{2\pi}{n}+i\sin \frac{2\pi}{n}\]
I will now state two elementary properties (without proof) of the complex roots of unity that I will make use of later:
\[\sum_{p=0}^{n-1} (\omega_n^p)^k= \begin{cases} n & \text{if} \space\space\space n|k\\ 0 & \text{else} \end{cases}\tag 1\]
\[\sum_{k=0}^{n-1} (\omega_n^p)^k= \begin{cases} n & \text{if} \space\space\space \omega_n^p=1\\ 0 & \text{else}\\ \end{cases}\tag 2\]

As a result,

Theorem 3:
If
\[G(x):=\sum_{k=1}^\infty x^k a_k\]
Then
\[\sum_{p=0}^{n-1}(\omega_n^p)^m G(\omega_n^p)=n\sum_{k=1}^\infty a_{nk-m}\]

Proof: This theorem follows directly from properties (1) and (2) of the complex roots of unity. \(\blacksquare\)

Using the solution to the generalized recurrence and these identities, one may prove the following:

Theorem 4:
\[\sum_{k=1}^\infty \frac{(-1)^{nk}}{F_{nk-m}F_{nk+n-m}}=(-1)^{m-1}\bigg[\frac{\phi^{-1}}{F_n}-\frac{F_{n-m-1}}{F_n F_{n-m}}\bigg],\space\space\space n\ge 2, \space\space 0\le m\le n-1\]

Proof: I shall divide this proof into two cases - one for even \(n\), and one for odd \(n\). We shall consider even \(n\) first.

Suppose \(n\) is even. I will begin with the following formula from Theorem 2:
\[\Phi_n(x)=\frac{\Phi_1(x)}{F_n}\sum_{k=0}^{n-1} (-x)^{-k}-\frac{1}{F_n}\sum_{k=0}^{n-1}\frac{(-x)^{1-k}F_{n-k-1}}{F_{n-k}}\]
Letting \(x=\omega_n^p\), we have
\[\Phi_n(\omega_n^p)=\frac{\Phi_1(\omega_n^p)}{F_n}\sum_{k=0}^{n-1} (-\omega_n^p)^{-k}-\frac{1}{F_n}\sum_{k=0}^{n-1}\frac{(-\omega_n^p)^{1-k}F_{n-k-1}}{F_{n-k}}\]
\[(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{(-1)^{m-1}\Phi_1(\omega_n^p)}{F_n}\sum_{k=0}^{n-1} (-\omega_n^p)^{m-k-1}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(-\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
By property (2) of the roots of unity, when \(\omega_n^p\ne -1\), this is equal to
\[(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(-\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
and when \(\omega_n^p=-1\), it is equal to
\[(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\Phi_1(-1)}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(-\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
\[(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(-\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
This tells us that, since exactly one of the \(\omega_n^p\) with \(p\) between \(0\) and \(n-1\) is equal to \(-1\),
\[\sum_{p=0}^{n-1}(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\sum_{p=0}^{n-1}\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(-\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
\[\sum_{p=0}^{n-1}(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\bigg(\frac{F_{n-k-1}}{F_{n-k}}\sum_{p=0}^{n-1}(-\omega_n^p)^{m-k}\bigg)\]
Now note that since the sum \(\sum_{p=0}^{n-1}(-\omega_n^p)^{m-k}\) cycles through the nth roots of unity, and since when \(n\) is even, \(\omega_n^p\) being a root of unity implies that \(-\omega_n^p\) is also a root of unity, it is equal to the sum \(\sum_{p=0}^{n-1}(\omega_n^p)^{m-k}\). Thus,
\[\sum_{p=0}^{n-1}(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\bigg(\frac{F_{n-k-1}}{F_{n-k}}\sum_{p=0}^{n-1}(\omega_n^p)^{m-k}\bigg)\]
and, by identity (1), this is
\[\sum_{p=0}^{n-1}(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\bigg(\frac{F_{n-k-1}}{F_{n-k}}\cdot\begin{cases} n & \text{if} \space\space\space n|m-k\\ 0 & \text{else} \end{cases}\bigg)\]
Now, if one assumes that \(0\le m\le n-1\), the only value of \(k\) between \(0\) and \(n-1\) satisfying \(n|m-k\) would be \(k=m\). Thus, all terms of the sum are multiplied by zero except for the \(k=m\) term, and all that remains is
\[\sum_{p=0}^{n-1}(\omega_n^p)^{m-1}\Phi_n(\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{n(-1)^{m-1}F_{n-m-1}}{F_nF_{n-m}}\]
Using Theorem 3, one may replace the LHS:
\[\sum_{k=1}^\infty \frac{n}{F_{nk-m}F_{nk+n-m}}=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{n(-1)^{m-1}F_{n-m-1}}{F_nF_{n-m}}\]
\[\sum_{k=1}^\infty \frac{1}{F_{nk-m}F_{nk+n-m}}=(-1)^{m-1}\bigg[\frac{\phi^{-1}}{F_n}-\frac{F_{n-m-1}}{F_nF_{n-m}}\bigg]\]
and, because \(n\) is even, this is equivalent to
\[\sum_{k=1}^\infty \frac{(-1)^{nk}}{F_{nk-m}F_{nk+n-m}}=(-1)^{m-1}\bigg[\frac{\phi^{-1}}{F_n}-\frac{F_{n-m-1}}{F_nF_{n-m}}\bigg]\]
Which completes the proof for the case of even \(n\).

Now suppose that \(n\) is odd. Again, I will begin with the following formula from Theorem 2:
\[\Phi_n(x)=\frac{\Phi_1(x)}{F_n}\sum_{k=0}^{n-1} (-x)^{-k}-\frac{1}{F_n}\sum_{k=0}^{n-1}\frac{(-x)^{1-k}F_{n-k-1}}{F_{n-k}}\]
but this time, I will let \(x=-\omega_n^p\), so that
\[\Phi_n(-\omega_n^p)=\frac{\Phi_1(-\omega_n^p)}{F_n}\sum_{k=0}^{n-1} (\omega_n^p)^{-k}-\frac{1}{F_n}\sum_{k=0}^{n-1}\frac{(\omega_n^p)^{1-k}F_{n-k-1}}{F_{n-k}}\]
\[(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{(-1)^{m-1}\Phi_1(-\omega_n^p)}{F_n}\sum_{k=0}^{n-1} (\omega_n^p)^{m-k-1}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
By property (2) of the roots of unity, when \(\omega_n^p\ne 1\), this is equal to
\[(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
and when \(\omega_n^p=1\), it is equal to
\[(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{n(-1)^{m-1}\Phi_1(-1)}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
\[(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
Because of this, one may say that
\[\sum_{p=0}^{n-1}(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\sum_{p=0}^{n-1}\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\frac{(\omega_n^p)^{m-k}F_{n-k-1}}{F_{n-k}}\]
\[\sum_{p=0}^{n-1}(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\bigg(\frac{F_{n-k-1}}{F_{n-k}}\sum_{p=0}^{n-1}(\omega_n^p)^{m-k} \bigg)\]
By property (1), this is
\[\sum_{p=0}^{n-1}(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\sum_{k=0}^{n-1}\bigg(\frac{F_{n-k-1}}{F_{n-k}}\cdot \begin{cases} n & \text{if} \space\space\space n|m-k\\ 0 & \text{else} \end{cases} \bigg)\]
If one assumes that \(0\le m\le n-1\), the only term of the sum that does not vanish is the term for which \(k=m\). Thus,
\[\sum_{p=0}^{n-1}(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=\frac{n(-1)^{m-1}\phi^{-1}}{F_n}-\frac{(-1)^{m-1}}{F_n}\frac{nF_{n-m-1}}{F_{n-m}}\]
\[\sum_{p=0}^{n-1}(-\omega_n^p)^{m-1}\Phi_n(-\omega_n^p)=(-1)^{m-1}\bigg[\frac{n\phi^{-1}}{F_n}-\frac{nF_{n-m-1}}{F_nF_{n-m}}\bigg]\]
By Theorem 3, the LHS can be replaced with an infinite series:
\[n\sum_{k=1}^\infty \frac{(-1)^{nk}}{F_{nk-m}F_{nk+n-m}}=(-1)^{m-1}\bigg[\frac{n\phi^{-1}}{F_n}-\frac{nF_{n-m-1}}{F_nF_{n-m}}\bigg]\]
and so, at last,
\[\sum_{k=1}^\infty \frac{(-1)^{nk}}{F_{nk-m}F_{nk+n-m}}=(-1)^{m-1}\bigg[\frac{\phi^{-1}}{F_n}-\frac{F_{n-m-1}}{F_nF_{n-m}}\bigg]\]
Which completes the proof for the case of odd \(n\). Thus I have established that
\[\sum_{k=1}^\infty \frac{(-1)^{nk}}{F_{nk-m}F_{nk+n-m}}=(-1)^{m-1}\bigg[\frac{\phi^{-1}}{F_n}-\frac{F_{n-m-1}}{F_nF_{n-m}}\bigg],\space\space\space n\ge 2, \space\space 0\le m\le n-1 \space\space\space\blacksquare\]

With this formula, the third and fourth proposed problems may be solved readily:
\[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{F_{5k-1}F_{5k+4}}= \frac{2}{15}-\frac{\phi^{-1}}{5}\]
\[\sum_{k=1}^\infty \frac{1}{F_{10k}F_{10k+10}}=\frac{\phi^{-1}}{55}-\frac{34}{3025}\]