Consider the following summation:
$$S_n:=\sum_{k=1}^{n}\frac{1}{k^{F_{k}}}$$
Where $F_n$ is the nth Fibonacci number.
Some of the values are listed below:
$$ \begin{array}{r|rr} n&S_n&\\\hline1&1\\ 5&\;\;\;1.627056\overline{1}&\\ 8&≈1.62705670649561296431951406885467949\\ 15&≈1.62705670649561296431951406885468308&\\ 20&≈1.62705670649561296431951406885468308&\\ 50&≈1.62705670649561296431951406885468308 \\ 100&≈1.62705670649561296431951406885468308\\ 200&≈1.62705670649561296431951406885468308\\ \end{array} $$
As it's clear the convergence of this sum is so fast.
Using the fact $k\le F_{k}$ for $6\le k$ we conclude that:
$$1<\sum_{k=1}^{\infty}\frac{1}{k^{F_{k}}}<\sum_{k=1}^{\infty}\frac{1}{k^{k}}<\sum_{k=1}^{\infty}\frac{1}{k!}=e-1≈1.71828182846$$
The question is that how it can be shown that what does the series converge to by calculation.