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I want to find an sharp upper bound for $$\sum_{n=1}^{\infty}\frac{1}{F_n}$$which $F_n $ is the n$th$ term of Fibonacci sequence . I wrote a Matlab program to find an upper bound ,$\sum_{n=1}^{10^6}\frac{1}{F_n}<4$
Now my question is:(1):Is there an inequality to find this ?
(2): Is that series have a close form ? $${F_n} = \frac{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }}\to \\\sum_{n=1}^{\infty}\frac{1}{F_n}=\sum_{n=1}^{\infty}\frac{\sqrt 5}{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}\\\leq \sum_{n=1}^{\infty}\frac{\sqrt 5}{{{(\frac{{1 + \sqrt 5 }}{2} ) ^n} }}=\frac{\sqrt5}{1-\frac{1}{\frac{{1 + \sqrt 5 }}{2}}}\approx12.18\\$$I am thankful for a hint or solution which can bring a sharper upper bound .

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  • $\begingroup$ maybe related $\endgroup$ – MAN-MADE Sep 1 '17 at 7:33
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    $\begingroup$ Simple fast program: 3.359885666243177553172011302918927179688905133731968486495553815325130318996683383615416216456790087297045342928853913304136789017100883679591351733077119078580333550332507753187599850487179777897006039564509215375892775265673354024033169441799293934610992626257964647651868659449710216558984360881472693249591079473873673378523326877499762727757946853676918541981467668742998767382096913901217722024405208151094264934951 $\endgroup$ – Kenny Lau Sep 1 '17 at 7:33
  • $\begingroup$ WolframAlpha simply returns ℱ suggesting that it has no closed form. $\endgroup$ – Kenny Lau Sep 1 '17 at 7:35
  • $\begingroup$ @KennyLau:Can I see your program ? can you post it here ? $\endgroup$ – Khosrotash Sep 1 '17 at 7:35
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    $\begingroup$ $F_n-\phi^n/\sqrt5=-(-1/\phi)^n/\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign, and that your inequality follows? $\endgroup$ – Jyrki Lahtonen Sep 1 '17 at 21:02
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As Jyrki Lahtonen points out in the comments, $\sum_{n=1}^\infty\frac{\sqrt5}{\varphi^n}$ isn't necessarily the right bound, since it doesn't dominate the original series term-by-term.

As for a closed form, this identity is known: $$\sum_{n=1}^\infty\frac{1}{F_n}=\frac{\sqrt5}{4}\left(\vartheta_2^2(\varphi^{-2}) + \frac{\log5 + 2\psi_{\varphi^{-4}}(1) - 4\psi_{\varphi^{-2}}(1)}{2\log\varphi}\right)$$ where $\vartheta_2(q)$ is the Jacobi theta function at $z=0$, and $\psi_q$ is the $q$-digamma function. See Wikipedia and MathWorld on the "Reciprocal Fibonacci constant", and other Math.SE questions such as What is the sum of Fibonacci reciprocals?

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    $\begingroup$ $F_n-\phi^n/\sqrt5=-(-1/\phi)^n/\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign? The OP suffers from the same problem. $\endgroup$ – Jyrki Lahtonen Sep 1 '17 at 20:59
  • $\begingroup$ @JyrkiLahtonen Whoops! Let me fix that. $\endgroup$ – Chris Culter Sep 1 '17 at 21:12
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Here is a systematic method for computing upper bounds for the sum without much work.

By induction, if $F_N \ge b^N$ and $b+1 \ge b^2$, then $F_n \ge b^n$ for $n \ge N$.

Therefore, $$ \sum_{n=1}^\infty {1\over F_n} \le \sum_{n=1}^{N-1} {1\over F_n} + \sum_{n=N}^\infty {1\over b^n} = \sum_{n=1}^{N-1} {1\over F_n} + \frac{1}{b^{N-1}(b-1)} $$ This upper bound gets closer to the actual sum when $b$ gets larger, but then we need larger $N$:

\begin{array}{crl} b &N &sum \\ 1.3 &4 &4.0172204521317 \\ 4/3 &5 &3.7825520833333 \\ 1.4 &6 &3.4981694135380 \\ 1.5 &11 &3.3651521005948 \\ 1.6 &72 &3.3598856662432 \\ \end{array}

Since $1.6 \approx \phi$, which is the largest possible $b$, the last value is quite close to the actual value: $$ 3.359885666243177553172011302918927179688905133732\cdots $$

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Using Wolfram Mathematica answer is : Wolfram Mathematica Code

enter image description here

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