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So, I want to find the closed value for the sequence $$\sum_{n=1}^{\infty} \frac{1}{F(n)F(n+2k)}$$ Here by $F(n)$ I mean the $n$-th term of Fibonacci Sequence.

I got motivated for this by the YouTube video "Using partial fractions to evaluate two Fibonacci reciprocal sums" by Prof. Michael Penn. So in the end of video he says for finding this for the general case : k. So i tried it and here is what i did:-

Let $$\frac{1}{F(n)F(n+2k)} = \frac{A}{F(n)} + \frac{B}{F(n+2k)}$$

so by rearranging we can say that: $AF(n+2k) + BF(n) = 1$

Now let $C_0 = F(n+2k)$

then $C_1 = F(n+2k-1) + F(n+2k-2)$

so $C_2 = 2F(n+2k-2) + F(n+2k-3)$

then by induction we can prove that if i repeat this process j times then $C_j = F(j+1)F(n+2k-j) + F(j)F(n+2k-j-1)$

So now let $j = 2k - 1$ F then $C_{2k-1} = F(2k)F(n+1) + F(2k-1)F(n)$

now putting this in to the Eq, of A and B then $AF(2k)F(n+1) + AF(2k-1)F(n) + BF(n) = 1$

now by putting $F(n+1) = F(n+2) - F(n)$

then I will get $AF(2k)F(n+2) + (B+AF(2K-1) - AF(2k))F(n = 1)$

there can be infinite values of A and B satisfying this relation so if I let $A = \frac{1}{F(2k)F(n+2)}$ becaues the first term in the Eq. will become 1 then $B = \frac{F(2k-2)}{F(2k)F(n+2)}$

By putting this values in the summation I will get $\sum_{n=1}^{\infty}\frac{1}{F(2k)F(n)F(n+2)} + \frac{F(2k-2)}{F(n+2)F(n+2k)}$

now in the video It is showned that $\sum_{n=1}^{\infty}\frac{1}{F(n)F(n+2)} = 1$ so I am not going to derive it.

Now putting it we get $\frac{1 + F(2k-2)\sum_{n=1}^{\infty}\frac{1}{F(n+2)F(n+2k)}}{F(2k)}$

$\frac{1 + F(2k-2)\sum_{n=3}^{\infty}\frac{1}{F(n)F(n+2(k-1))}}{F(2k)}$ in this I have changed the limit.

Now by adding and subtracting some termes I get $Sum(k) = \frac{1 + F(2k-2)\sum_{n=1}^{\infty}\frac{1}{F(n)F(n+2(k-1))} - \frac{F(2k-2)}{F(2k-1)} - \frac{F(2k-2)}{F(2k)}}{F(2k)}$

Now finnaly I ended up with this $Sum(k) = \frac{1 + F(2k-2)Sum(k-1) - \frac{F(2k-2)}{F(2k-1)} - \frac{F(2k-2)}{F(2k)}}{F(2k)}$

But this is not a direct formulae. Can someone help?

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After revising your work, you should find $$\sum_{n=1}^{\infty} \frac{1}{F_nF_{n+2k}}=\frac{1}{F_{2k}}\sum_{j=1}^k\frac{1}{F_{2j-1}F_{2j}}.$$ The first terms for $k\geq 1$ are $1$, $\frac{7}{18}$, $\frac{143}{960}$, $\frac{4351}{76440}$, $\frac{814001}{37437400}$ (see also OEIS sequence A333088).

Proof. By induction with respect to $k\geq 1$, we verify that $$\frac{F_{2k}}{F_nF_{n+2k}}=\sum_{j=1}^k\frac{1}{F_{n+2(j-1)}F_{n+2j}}.$$ Indeed, it holds for $k=1$ and, for the inductive step, we have $$\sum_{j=1}^{k+1}\frac{1}{F_{n+2(j-1)}F_{n+2j}} =\frac{F_{2k}}{F_nF_{n+2k}}+\frac{1}{F_{n+2k}F_{n+2k+2}}=\frac{F_{2k+2}}{F_nF_{n+2k+2}}.$$ where at the last step we applied Vajda's identity: $$F_{2k+2}F_{n+2k}-F_{2k}F_{n+2k+2}=(-1)^{2k}F_2F_n=F_n.$$

Hence, since $\sum_{n=1}^{\infty} \frac{1}{F_nF_{n+2}}=1$ (watch the video or see Sum $\frac{1}{1\times2}+\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times8}+\cdots$), it follows that \begin{align} \sum_{n=1}^{\infty} \frac{1}{F_nF_{n+2k}}&= \frac{1}{F_{2k}}\sum_{n=1}^{\infty}\sum_{j=1}^k\frac{1}{F_{n+2(j-1)}F_{n+2j}}\\ &=\frac{1}{F_{2k}}\sum_{j=1}^k\sum_{n=1}^{\infty}\frac{1}{F_{n+2(j-1)}F_{n+2j}}=\frac{1}{F_{2k}}\sum_{j=1}^k\left(1-\sum_{n=1}^{2(j-1)}\frac{1}{F_{n}F_{n+2}}\right)\\ &=\frac{1}{F_{2k}}\sum_{j=1}^k\left(1-\sum_{n=1}^{2(j-1)}\left(\frac{1}{F_{n}F_{n+1}}-\frac{1}{F_{n+1}F_{n+2}}\right)\right)\\ &=\frac{1}{F_{2k}}\sum_{j=1}^k\left(1-1+\frac{1}{F_{2j-1}F_{2j}}\right) =\frac{1}{F_{2k}}\sum_{j=1}^k\frac{1}{F_{2j-1}F_{2j}}. \end{align}

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  • $\begingroup$ How did you reached there ? $\endgroup$ – Dhaval Bothra Aug 10 '20 at 17:01
  • $\begingroup$ @DhavalBothra Is it clear now? $\endgroup$ – Robert Z Aug 10 '20 at 17:16
  • $\begingroup$ yes thank btw can there be an anoather method and sorry if in the question i made any mistake because I am of 13 years. $\endgroup$ – Dhaval Bothra Aug 10 '20 at 17:19
  • $\begingroup$ @DhavalBothra I don't know... I followed the approach provided in the video. $\endgroup$ – Robert Z Aug 10 '20 at 17:23
  • $\begingroup$ Just elaborate the part that You have written in We have that $\endgroup$ – Dhaval Bothra Aug 11 '20 at 4:35
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty} {1 \over \mrm{F}\pars{n}\mrm{F}\pars{n + 2k}}} = \sum_{n = 1}^{\infty}{1 \over \mrm{F}^{\, 2}\pars{n + k} - \bracks{\ic^{n}\,\mrm{F}\pars{k}}^{2}} \\[5mm] = &\ {1 \over 2\,\mrm{F}\pars{k}}\sum_{n = 1}^{\infty} \bracks{% {\pars{-\ic}^{n} \over \mrm{F}\pars{n + k} - \ic^{n}\,\mrm{F}\pars{k}} - {\pars{-\ic}^{n} \over \mrm{F}\pars{n + k} + \ic^{n}\,\mrm{F}\pars{k}}} \\[5mm] = &\ {\mathcal{F}\pars{\mrm{F}\pars{k}/2} - \mathcal{F}\pars{-\mrm{F}\pars{k}/2} \over 2\,\mrm{F}\pars{k}} \quad\mbox{where}\quad \mathcal{F}\pars{z} \equiv \sum_{n = 1}^{\infty} {\pars{-\ic}^{n} \over \mrm{F}\pars{n + k} - 2 \ic^{n}\, z} \end{align}


\begin{align} \mathcal{F}\pars{z} & \equiv \sum_{n = 1}^{\infty} {\pars{-\ic}^{n} \over \mrm{F}\pars{n + k} - 2\ic^{n}\, z} = \sum_{n = 1}^{\infty}{\pars{-\ic}^{n} \over \phi^{n + k} - \pars{-1}^{n + k +1}\,\phi^{-n - k} - 2\ic^{n}\, z} \\[5mm] & = \sum_{n = 1}^{\infty}{\pars{-\ic}^{n}\phi^{n + k} \over \phi^{2n + 2k} - 2\ic^{n}z\,\phi^{n + k} - \pars{-1}^{n + k +1}} = \sum_{n = 1}^{\infty}{\pars{-\ic}^{n}\phi^{n + k} \over \pars{\phi^{n + k} - \ic^{n}\, r_{-}}\pars{\phi^{n + k} - \ic^{n}\, r_{+}}} \end{align} $\ds{\ic^{n}\, r_{\pm}}$ are the roots of $\ds{r^{2} -2\ic zr - \pars{-1}^{n + k + 1} = 0}$. Namely, $$ r_{\pm} = \pars{z \pm \Delta}\,,\qquad \Delta \equiv \root{z + \pars{-1}^{k + 1}} $$ Then, \begin{align} \mathcal{F}\pars{z} & = {1 \over 2}\pars{1 - {z \over \Delta}}\sum_{n = 1}^{\infty}{\pars{-\ic}^{n} \over \phi^{n + k} - \ic^{n}\, r_{-}} + {1 \over 2}\pars{1 + {z \over \Delta}}\sum_{n = 1}^{\infty}{\pars{-\ic}^{n} \over \phi^{n + k} - \ic^{n}\, r_{+}} \end{align} So, we are left with evaluation of a sum likes $\ds{\bbox[10px,#ffd]{\left.\sum_{n = 1}^{\infty}{\pars{-\ic}^{n} \over \phi^{n + k} - \ic^{n}\, \xi}\,\right\vert_{\large\ \xi\ \in\ \mathbb{C}}}}$
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  • $\begingroup$ does i means root(-1) in the answer $\endgroup$ – Dhaval Bothra Aug 11 '20 at 7:25
  • $\begingroup$ Yes. Thanks. ${}{}{}{}{}{}$ $\endgroup$ – Felix Marin Aug 11 '20 at 12:14
  • $\begingroup$ I have not accepted it beacuse it is difficult to understand . But yes your answer is really good $\endgroup$ – Dhaval Bothra Aug 11 '20 at 15:32

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