Questions are about this paper: http://users.uoi.gr/abeligia/gorenstein.pdf and all algebras are finite dimensional

Question 1: In corollary 6.21 (2) there is a proof of the direction (e) implies (a): It is said that if an algebra has finite finitistic dimension d, then $\Omega^d(mod-A)=\Omega^{d+i}(mod-A)$ for all $i \geq 0$ (a paper of auslander reiten is quoted, but I cant find the exact statement there, there is however a similar statement with the additional assumption that $\Omega^i(mod-A)$ is extension closed for all i).

So is the assertion of this proof in the paper correct?

Here an (possible counter) example: Let A be the Nakayama algebra with Kupisch series [3,4]. This algebra has finitistic dimension 1 and I think the first simple module is in $\Omega^{1}(mod-A)$ but not in $\Omega^2(mod-A)$, is this correct?

Question 2: In the same place in the paper the author claims that (a) implies for example (b), so that any Gorenstein algebra should be k-Gorenstein for all k. But on page 12 on the top there are counterexamples http://ac.els-cdn.com/0022404994900442/1-s2.0-0022404994900442-main.pdf?_tid=8201a554-7aab-11e7-87c3-00000aab0f6c&acdnat=1502026449_b3cbd17ed58cec1c97d44c6e3d508e59 . Do I confuse something?

Question 3: Are there easy example of algebras $A$ such that $\Omega^i(mod-A) \subset \Omega^{i-1}(mod-A)$ strictly for every i? Note that such algebras have to be a little exotice since $\Omega^i(mod-A)$ cant be representation-finite for any i (this excludes for example representation-finite or monomial algebras). I have many candidates for such algebras (including algebras which might have infinite finitstic dimension :) ) but it seems to be not so easy to prove it.

To question 1: I used the following in my calculations to test wheter an indecomposable non-projective module M is in $\Omega^i(mod-A)$: $M$ is in $\Omega^i(mod-A)$ iff $M$ is a summand of $\Omega^{i}(\Omega^{-i}(M))$, using https://folk.ntnu.no/oyvinso/Papers/rel1.pdf proposition 3.2. Hope this is correct.

edit: Assume an algebras is nearly Gorenstein (meaning that Gorenstein projectives are equal to $^{\perp}A$ and the dual property). For example all representation-finite algebras are nearly Gorenstein. Then the condition for being Co-Gorenstein is $\Omega^{\infty} \subseteq ^{\perp}A=Gp(A)$, but always $Gp(A) \subseteq \Omega^{\infty}$, so the condition is $Gp(A) = \Omega^{\infty}$. Now if the algebra is $\Omega^{r}$-finite for some $r$ (as representation-finite algebras are), then $\Omega^{\infty}=\Omega^{s}$ for some s and thus the condition is $\Omega^{s}=Gp(A)$, but this is equivalent to $A$ being s-Gorenstein. Thus in this situation Co-Gorenstein implies Gorenstien. This applies for example to torsionless-finite algebras (which include all representation-finite algebras).

edit2: Here an example of a Gorenstein algebra (thus CoGorenstein algebra) with $\Omega^{i}$ not extension closed for all i: Take the Nakayama algebra with Kupisch series [2,3,3,2,1]. By the main result of the paper "Syzygy modules for noetherian rings" by Auslander and Reiten, $\Omega^i$ is extension closed for $1 \leq i \leq k$ iff the projective dimesnion of $I_i$ is bounded by $i+1$ for all $i < k$. Now for the nakayama algebra corresponding to [2,3,3,2,1] with minimal projective resolution $P_i$ of D(A), the module $P_1$ has injective dimension 3 and 3>2 thus the condition is not satisfied.

rightmodule. This means that your example should be the algebra with Kupisch series $[2,3,3,2,1]$ and not its opposite algebra. Even without the Auslander-Reiten result it's easy to check directly that $\Omega^2$ is not closed under extensions (but it is for the opposite algebra, which is what was confusing me). – Jeremy Rickard Aug 9 '17 at 9:27