How to prove that the series $\sum _{n=1}^{\infty } \left( {F_{n+1}} \right) ^{- {F_n}} \approx 1.619141630$

Question

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.

Answer 1

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).