The current best bound for the maximum possible density of an $n$-node graph with girth (shortest cycle length) $>2k$ is of the form
$$ex(n \ \mid \ C\_{\le 2k}) = O(n^{1 + 1/k}),$$
while the current best bound for the maximum possible density of a graph without a $2k$-cycle is of the form
$$ex(n \ \mid C\_{2k}) = O(k \cdot n^{1 + 1/k}).$$
[Edit: this bound is actually wrong, see accepted answer below.]
An "intermediate" question is the maximum possible density of a graph without a "non-backtracking $2k$-cycle," which we define as a circularly-ordered sequence of $2k$ nodes in which adjacent nodes are connected by an edge and there is no continuous subsequence of the form $(u, v, u)$. It is not hard to see that this function $ex(n \ \mid \ C^{\not \leftarrow}\_{2k})$ lies between the above two, but:



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> 1. Is $ex(n \ \mid \ C^{\not \leftarrow}\_{2k})$ known to be asymptotically equivalent to either of these two functions?
> 2. (If not) is any upper bound on $ex(n \ \mid \ C^{\not \leftarrow}\_{2k})$ known that is asymptotically better than $O(k \cdot n^{1 + 1/k})$?
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Extremal density of a graph without a non-backtracking $2k$-cycle
The best bounds on $ex(n, C^{\not \leftarrow}\_{2k})$ and on $ex(n, C\_{2k})$ that we can prove with the current techniques are going to be basically same. Below I explain why.


First, contrary to what the question states the best known bound on the number of edges in $C\_{2k}$-free graphs is $O(\sqrt{k}\log k\cdot n^{1+1/k})$, as proved in [a paper of Jiang and myself](http://arxiv.org/abs/1403.1601). 


In the appendix of that paper, we also show that this bound is almost the best one can obtain without improving the bound on $ex(n,C\_{\leq 2k})$ itself. That construction also works for non-backtracking cycles. For completeness, I review the construction below.


For simplicity assume that $k$ is even. Consider bipartite version of the girth problem. Namely, we seek a $C\_{\leq 2k}$-free bipartite graph whose parts have size $m$ and $n$. The same argument as in non-bipartite case shows that the number of edges in this case is at most $cn^{1/k}(mn)^{1/2}$ if $k$ is even. Let $m=n/k$ and suppose that this bound is tight, i.e., there is $C\_{\leq 2k}$-free bipartite graph with parts $n/k$ and $n$ with $Ck^{-1/2}n^{1+1/k}$ edges. Replace each vertex in the smaller part by an independent set of size $k$ to get a graph $G$ with $Ck^{1/2}n^{1+1/k}$ edges. This graph contains no $C^{\not \leftarrow}\_{2k}$ as the vertices of $C^{\not \leftarrow}\_{2k}$ would induce a cycle length at most $2k$ in the original graph.


So, apart from the $\log k$ factor, no improvement is going to happen until we improve on the basic girth argument.


