What is the maximum number of edges in a $k$-connected (but not $(k+1)$-connected) $n$-vertex graph for $n \geq 6$ and $k = 2,3$? My approach for finding the $\textbf{minimum}$ number of edges was to use that $\max k \leq $ min degree $ \leq \frac{2e}{n} $. I do not know if a similar idea would help for the maximum.
(A graph $G$ is $k$-connected if it has at least $k+1$ vertices and no matter which $k-1$ vertices we delete (along with their edges), the graph remains connected.)
Any help appreciated!
Now I'm sitting at a graph theory conference, so I post this draft answer, which I'm going to finish later. :-)
Let $G$ be such a graph. Since $G$ is not $k+1$, we can delete $k$ vertices of $G$ such that the residual graph splits into at least two connected components of sizes $n_1,\dots, n_l$. Then they have at total at most
$$\sum {n_i\choose 2}=\frac 12\sum n_i^2-n_i=\frac 12\left(k-n+\sum n_i^2\right).$$
An inequality $r^2+s^2\le (r-1)^2+(s+1)^2$, valid for all non-negative $r\le s+1$, implies that the maximum $M$ of the sum $\sum n_i^2$ is attained when one of $n_i$ is $n-k-1$, other is $1$, and the remaining are zeroes. That is $M=(n-k-1)^2+1^2$. The removed $k$ vertices can contribute at most ${k\choose 2}+k(n-k)$ edges to $G$. That is, the graph $G$ has at most
$$(n-k-1)^2+1+{k\choose 2}+k(n-k)=...$$ edges.
On the other hand, the suggested by these estimation extremal graph should be $k$-connected.