Does k moments imply k + $\epsilon$ moments? If you have a random variable $X$ such that $\mathbb{E}(|X|^k) < \infty$, does it follow that  $\mathbb{E}(|X|^{k+\epsilon}) < \infty$ for some (potentially small) $\epsilon > 0$? If not, what is a good counterexample? I.e. a random variable such that $\mathbb{E}(|X|^k) < \infty$ but $\mathbb{E}(|X|^{k+\epsilon}) = \infty$ for all $\epsilon > 0$.
I know that in general $\mathbb{E}(|X|^{k+\delta}) < \infty$ some $\delta > 0$ would imply that $\mathbb{E}(|X|^k) < \infty$, but I have not been able to find e.g. an example of a random variable which only has second moments (for instance) and not anything higher at all.
I do know that it is possible to have everything up to k moments and not have the kth moment (see e.g. the example in the first answer here: https://math.stackexchange.com/questions/1955968/example-of-random-variable-that-is-integrable-but-have-infinite-second-moment).

It suffices to find a counter-example for $k=1$, that is, a random variable $Y\geqslant 0$ such that $\mathbb E[Y]$ is finite but $\mathbb E\left[Y^{1+\varepsilon}\right]$ is infinite and $X=Y^k$ will give the wanted counter-example.
Let $c=\sum_{j=2}^\infty \left(j \log j\right)^{-2}$ and let $p_j=c^{-1} \left(j \log j\right)^{-2}$, so that $\sum_{j\geqslant 2}p_j=1$. Let $Y$ be a random variable taking the value $j$ (for $j\geqslant 2$) with probability $p_j$. Then
$$
\mathbb E[Y]=\sum_{j=2}^\infty j\mathbb P(Y=j)=c^{-1}\sum_{j=2}^\infty\frac 1{j(\log j)^2}<\infty
$$
and for $\varepsilon>0$,
$$
\mathbb E\left[Y^{1+\varepsilon}\right]=\sum_{j=2}^\infty j^{1+\varepsilon}\mathbb P(Y=j)=c^{-1}\sum_{j=2}^\infty\frac 1{j^{1-\varepsilon}(\log j)^2}=\infty.
$$
