We'll use that any negligible function converges to $0$ as it's input goes to $+\infty$ (which follows from the definition of a negligible function). Notice that the converse does not hold.
Does overwhelming imply non-negligible ? Equivalently: Is it necessarily true that $f$ is not-negligible if $1−f$ is negligible?
Yes. Proof: If $1−f$ is negligible, then $1-f(n)$ converges to $0$ as $n$ goes to $+\infty$, thus $f(n)$ converges to $1$ as $n$ goes to $+\infty$. Thus $f$ can't be negligible, because a function can't converge to two different values when it's input goes to $+\infty$.
(if $1−f$ is negligible then) $f$ must get arbitrarily close to $1$ at some exponential rate.
No. A counterexample is $f(n)=1-1/n^{\ln n}$. It's such that $1-f(n)=1/n^{\ln n}$, which is negligible, but does not converge to $0$ at an exponential rate as $n$ goes to $+\infty$. It converges to $0$ faster than any polynomial, but slower than exponentially.
Note: functions like $f(n)=n^{-\ln(\ln n)/100}$, which are theoretically negligible but converge so slowly for practical bit size $n$, illustrate there are limitations to theoretical notions of security relying on the theoretical definition of "negligible". Practitioners must check that practical parameters are such that the probability of success of an actual adversary is actually negligible, which depending on context can be $<2^{-30}$ (one in a billion) or even much lower.
