What can be said about $\sum a_n^{\frac{n-1}{n}} $?

Hint: Let \begin{align} A &= \{n\in\mathbb{N}: a_n<2^{-n}\}\\ B &= \{n\in\mathbb{N}: a_n\ge 2^{-n}\} \end{align} (notice that $A\cup B = \mathbb{N}$). Then $$\sum\limits_{n\in A}{a_n^{\frac{n-1}{n}}}\le\sum\limits_{n\in A}{2^{-(n-1)}}$$ while for $n\in B$, we have $a_n\ge 2^{-n}\implies a_n^{-1/n}\le 2$, and hence $$\sum\limits_{n\in B}{a_n^{\frac{n-1}{n}}}\le\sum\limits_{n\in B}{2a_n}.$$