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If $F(n)$ is the nth Fibonacci number, How can I prove that: $$\sum_{i=1}^{\infty} \frac{1}{F(i)}\approx 3.36\, .$$

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$$ \begin{align} F_n &=\frac{\phi^n-(-1/\phi)^n}{\sqrt5}\\ &=\frac{\phi^n}{\sqrt5}\left(1-\left(-\frac1{\phi^2}\right)^n\right) \end{align} $$ Therefore, $$ \begin{align} \sum_{n=1}^\infty\frac1{F_n} &=\sum_{n=1}^\infty\frac{\sqrt5}{\phi^n}\left(1+\left(-\frac1{\phi^2}\right)^n+\left(-\frac1{\phi^2}\right)^{2n}+\left(-\frac1{\phi^2}\right)^{3n}+\cdots\right)\\ &=\sqrt5\left(\frac1{\phi-1}-\frac1{\phi^3+1}+\frac1{\phi^5-1}-\cdots\right)\\ &=\sqrt5\sum_{k=0}^\infty\frac{(-1)^k}{\phi^{2k+1}-(-1)^k} \end{align} $$ Since $\frac{\sqrt5}{\phi^{19}+1}=0.0002392$, $$ \sqrt5\sum_{k=0}^8\frac{(-1)^k}{\phi^{2k+1}-(-1)^k}=3.3600587 $$ is less than $0.0002392$ too high.

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  • $\begingroup$ It is good approach for approximation of exact value of this summation. Thanks $\endgroup$ – Amin235 Mar 24 '16 at 9:46
  • $\begingroup$ +1 This is one of the good ways how to estimate the error. $\endgroup$ – yo' Mar 24 '16 at 10:01
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    $\begingroup$ @Amin23 : Yes. That bound is zero, uniformly in $n$. en.wikipedia.org/wiki/… $\endgroup$ – Eric Towers Mar 24 '16 at 12:42
  • $\begingroup$ Is there a bound or function based on parameter $n$ for this approximation $\mid F(n)-\frac{\phi^n-{(-1/\phi)}}{\sqrt{5}} \mid$ $\endgroup$ – Amin235 Mar 24 '16 at 12:47
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    $\begingroup$ As $n$ gets larger, that error tends to $\frac1{\phi\sqrt5}$, it does not vanish. $\endgroup$ – robjohn Mar 24 '16 at 14:34
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Since $F(n) \approx \frac{\phi^n}{\sqrt 5}$ for large enough $n$, you may use that as an approximation, which give you a geometric series. For instance, if we use the approximation from the fifth term on, we get $$ \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \sum_{n = 4}^\infty \frac{\sqrt 5}{\phi^n} \approx 3.3612 $$ To get a feel for how accurate this is, $F(5) = 5$, while $\frac{\phi^5}{\sqrt 5} \approx 4.96$, which is a relative error of less than $1\%$. The relative error gets smaller and smaller as $n$ grows, by a factor of about $3$, and it is alternating (every other term is too large, every other term is too small) which also reduces the error.

Since every term in the sum is less than $1\%$ away from the corresponding term in the original sequence, the true answer is within $1\%$ of the sum we have, which is $\sum_{n = 4}^\infty \frac{\sqrt 5}{\phi^n} \approx 0.53$. Thus the error of this approximation is at most about $0.005$. One more exact term, and the absolute error will be at most $0.001$.

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  • $\begingroup$ Thanks for useful comment. Is there a simple analytical method for proofing this summation? $\endgroup$ – Amin235 Mar 24 '16 at 7:52
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    $\begingroup$ @Amin23 What do you mean "proofing"? Do you mean calculating $\sum_{n = 5}^\infty \frac{\sqrt 5}{\phi^n}$? Then the answer is "yes", and the exact value is $\frac{\sqrt 5}{\phi^4(\phi - 1)}$. For more details, you can google "sum of convergent geometric series", it will tell you much more than I can tell you in a comment. $\endgroup$ – Arthur Mar 24 '16 at 7:57
  • $\begingroup$ @Jean-ClaudeArbaut The error is less that $1\%$, as I said in my answer (because every term is less than $1\%$ away from the corresponding term in the true sequence), so the absolute error is at most $0.005$. The error of the next term is about $0.3\%$, so if you want better accuracy you can just add more terms. $\endgroup$ – Arthur Mar 24 '16 at 8:02
  • $\begingroup$ @Arthur I find it it's analytical method in this site mathworld.wolfram.com/ReciprocalFibonacciConstant.html $\endgroup$ – Amin235 Mar 24 '16 at 8:04
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    $\begingroup$ @Amin23 So what you really want is not to prove that $\sum_{i = 1}^\infty 1/F(i) \approx 3.36$, but you rather want to calculate the sum $\sum_{i = 1}^\infty 1/F(i)$? Then I have no idea, and that page is probably your best bet. $\endgroup$ – Arthur Mar 24 '16 at 8:16

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