Here is a problem. Say $n= 2^k$ for some $k\in \mathbb{N}$. Find the smallest $\epsilon>0$ such that $$2\cdot 2^{n\over 2}+ 2^2\cdot 2^{n\over 4}+...+2^k\cdot 2^{n\over 2^k}<2^{n^{\epsilon}}$$ for sufficiently large $n$.
My work:
I managed to prove that $\epsilon =1$ works, but I don't know if it is the smallest one. It is obviously that $2^s\cdot 2^{n\over 2^s}<2^s\cdot 2^{n\over 2}$ for each $1<s\leq k$. If $f(n)$ is the sum then we have $$f(n) <2^{n\over 2} (2+2^2+...+2^k) = 2^{n\over 2} 2{2^k-1\over 2-1} = 2(n-1)\cdot 2^{n\over 2} $$ Since $$2(n-1)\cdot 2^{n\over 2} < 2^n$$ for $n\geq 8$ we are done.
So can we get $\epsilon <1$?
Notice that the sum is always greater than the first term, which is greater than $2^{\frac{n}2}.$ Notice that $\frac{n}{2} \gg n^\epsilon$ for any $\epsilon < 1,$ so your $\epsilon =1$ is the right answer.