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Suppose that $F_n$ is the $n$th term of Fibonacci numbers. By numerical calculation I see that $$ \sum _{n=1}^{\infty } \left( {F_{n+1}} \right) ^{- {F_n}} \approx 1.619141630 $$ The rate of convergence of the above series is too high. I mean, if we compute with 50 digits, for $n\geq 8$ the values of series is constant and is as follows $$ \sum _{n=1}^{n\geq8} \left( {F_{n+1}} \right) ^{- {F_n}} \approx 1.6191416299151308574250170831329152667545274408795 $$ Now my question: How to proof that the above series is converge to $1.619141630$.

Thanks for any suggestion.

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    $\begingroup$ These terms are going to zero very fast, all you need to do is show they are eventually bounded by a convergent geometric series. $\endgroup$ – Nate Jun 2 '17 at 22:33
  • $\begingroup$ Is the different indexing in the two displayed formulas (first base F_{n+1}, then F_n) intentional? $\endgroup$ – TMM Jun 2 '17 at 22:42
  • $\begingroup$ @TMM Is it possible to explain more your question. I can not figure out it. Thanks $\endgroup$ – Amin235 Jun 2 '17 at 22:46
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    $\begingroup$ First you wrote $(F_{n+1})^{-F_n}$ and then $(F_n)^{-F_n}$. Was that intentional, or is it a typo? $\endgroup$ – TMM Jun 2 '17 at 22:49
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    $\begingroup$ The Fibonacci numbers have a closed form, without recursion, derived here: {en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression} And the sum you are getting looks an awful lot like the Golden Ratio--see the link I cited.:) $\endgroup$ – avs Jun 3 '17 at 0:07
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It is easy to prove that $F_n>n$ for $n>5$.

Hence, $\left(F_{n+1}\right)^{-F_n}<n^{-n}<e^{-n}$ for $n>5$.

Let $$x=\sum_{n=1}^{\infty} \left(F_{n+1}\right)^{-F_n}.$$

Then, for $k$ an integer greater than $5$, $$x - \sum_{n=1}^{k} \left(F_{n+1}\right)^{-F_n} = \sum_{n=k+1}^{\infty} \left(F_{n+1}\right)^{-F_n} < \sum_{n=k+1}^{\infty} e^{-n} = \frac{e^{-k}}{e-1}$$

Choosing $k=23$, we can conclude that $$ 1.619141629915 < x < 1.619141629974$$ and so, with all digits correct, $x$ is approximately 1.6191416299.

You can certainly get more precision by using something smaller than $e^{-n}$ (as $n^{-n}$ is much smaller than that, but it's nice to have an easy tail).

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  • $\begingroup$ Thanks for your answer. I think the number $1.619141630$ has a closed form like the golden number ($\frac{1+\sqrt{5}}{2}$). $\endgroup$ – Amin235 Jun 3 '17 at 5:47
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    $\begingroup$ Why do you think that? Plugging this value into the inverse symbolic calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) yields nothing. $\endgroup$ – Matthew Conroy Jun 3 '17 at 5:59
  • $\begingroup$ In case anyone is curious, the continued fraction begins [1, 1, 1, 1, 1, 1, 2, 23, 1, 2, 1, 5, 7, 3, 2, 1, 3, 6, 15, 2, 1, 5, 2, 2, 1, 3, 2, 3, 3, 1, 14, 212, ...]. $\endgroup$ – Matthew Conroy Jun 3 '17 at 6:08
  • $\begingroup$ Just it was a guess. Is it true that we conclude your method is correct not only for Fibonacci numbers but also for any increasing sequence number? Thanks $\endgroup$ – Amin235 Jun 3 '17 at 15:20
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    $\begingroup$ Yes, this method (i.e., bounding the tail of a series to get approximations of the sum) applies to many, many sums (and integrals). $\endgroup$ – Matthew Conroy Jun 3 '17 at 17:53

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