Simple example of a distribution on +ve natural numbers with $\sum_{n=1}^\infty n^2 p_n < \infty$ and $\sum_{n=1}^\infty \sqrt{p_n} = \infty$

Question

DISCLAIMER: This is probably too elementary and I haven't spent much time thinking about it (as I should).

Question

What is a simple example of a distribution $(p_1,p_2,\ldots,p_n,\ldots)$ on +ve natural numbers (meaning that $p_n \ge 0\;\forall n =1,2,\ldots$ and $\sum_{n=1}^\infty p_n = 1$) satisfying simultaneously

• finite-second moment: $\sum_{n=1}^\infty n^2 p_n < \infty$
• divergent square root of probabilities: $\sum_{n=1}^\infty \sqrt{p_n} = \infty$ ?

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Edit

User Stephan Lafon has given a really simple proof which deserves to be expanded upon. In fact one can proof a stronger statement.

For any $\alpha \in (1, \infty)$, there is no distribution $p:=(p_1,\ldots,p_n,\ldots)$ with fintite $\alpha$-moment such that $\sum_n p_n^{1/\alpha} = \infty$.

Proof. Let $\beta := \alpha / (\alpha-1) > 1$ be the harmonic conjugate of $\alpha$. Define the functions $f,g: \{1,2,\ldots\} \rightarrow \mathbb R_+$ by $f(n) := 1/n$ and $g(n) := np_n^{1/\alpha}$. If $\sum_n n^\alpha p_n < \infty$, then one computes $$ \begin{split} \sum_n p_n^{1/\alpha} &= \sum_n (1/n) np^{1/\alpha} = \langle f, g\rangle \overset{\text{Cauchy-Schwarz}}{\le}\|f\|_\beta\|g\|_\alpha = \left(\sum_n 1/n^\beta\right)^{1/\beta}\left(\sum_n n^\alpha p_n\right)^{1/\alpha}\\ &= \zeta(\beta)^{1/\beta}\left(\sum_n n^\alpha p_n\right)^{1/\alpha} < \infty. \end{split} $$

Answer 1

There is no such distribution. Proof: By Cauchy-Schwarz: $$\sum_{n=1}^\infty \sqrt{p_n} = \sum_{n=1}^\infty \frac{n}{n}\sqrt{p_n} \leq \left(\sum_{n=1}^\infty n^2p_n\right)^{\frac1 2}\left(\sum_{n=1}^\infty\frac 1 {n^2}\right)^{\frac1 2}$$