TITLE: How to prove that the series $\sum _{n=1}^{\infty } \left( {F_{n+1}} \right) ^{- {F_n}} \approx 1.619141630$
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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).