How to construct a probability mass function which has a finite moment of order r but has no higher finite moment?
My approach:
I was trying to construct a probability mass function which is $ \propto \frac{1}{x^{r+2}}$, but I'm failing to construct one, since the sum $ \sum_{1}^{\infty}\frac{1}{x^{r+2}} $ is unsolvable (as far as I know).
Thanks in advance!
On
The reason your idea doesn't work is that $f(n)=\frac{n^{-p}}{\sum_n n^{-p}}$ has finite moments for $r<p-1$ but not equal to $p-1$.
A quick fix is to consider a sequence $p_k \to (r+1)^+$ and then sum up. That means taking $f(n)=\sum_{k=1}^\infty c_k n^{-p_k}$. This automatically doesn't have any moments higher than $r$ as long as all $c_k$ are positive.
To make this have a finite $r$th moment, it suffices to choose $c_k$ decaying fast enough. Specifically, notice that $\sum_{n=1}^\infty n^{r-p_k} \sim \int_1^\infty x^{r-p_k} dx = \frac{1}{p_k-(r+1)}$, since $p_k>r+1$ by assumption. So given a summable positive sequence $d_k$, you can take $c_k=\frac{d_k(p_k-(r+1))}{C}$. Then you choose $C$ for normalization.
The final result of this calculation is $f(n)=\frac{\sum_{k=1}^\infty d_k (p_k-(r+1)) n^{-p_k}}{\sum_{k=1}^\infty \sum_{n=1}^\infty d_k (p_k-(r+1)) n^{-p_k}}$. Here $p_k$ is any sequence converging to $r+1$ from the right, while $d_k$ is any summable positive sequence.
Adding in a power of $\ln(n)$ in the denominator is an easier fix, but arguably less in the spirit of your original idea.
Take the pmf as $\frac c {(n^{r+1}) (\ln n))^{2}}$ for $n\geq 2$ where $c$ is chosen such that the sum is $1$. I will let you verify that this has finite $r-$th moment but no higher moment.