Existence of sequence of distributions

Question

This question concerns distributions $\mu$ over the naturals $\mathbb{N}=\{1,2,\ldots\}$. For $q\ge1$, let us define the $q$th moment of entropy: $$ H_q(\mu)=\sum_{i=1}^\infty \mu(i)|\log\mu(i)|^q, $$ so $H_1(\mu)$ is just the usual entropy.

I am interested in a sequence of distributions $\mu_n$ satisfying the following properties:

1. $\mu_n(1)\to1$ as $n\to\infty$
2. $\limsup_{n\to\infty}H_2(\mu_n)<\infty$
3. $\liminf_{n\to\infty}H_1(\mu_n)>0$

Does such a sequence exist? I'd be satisfied with mere existence (though a construction would, of course, be nice).

If such a sequence does not exist (proof?), I'd be happy with the following weakening:

1'. $\mu_n(1)\to1$ at some fixed rate

2'. same as 2

3'. $H_1(\mu_n)\to0$ at an arbitrarily slow rate.

Answer 1

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

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After the editing of the question, the answer becomes no. Indeed, suppose that $\mu_n(1)=1-t_n$, where $t_n\downarrow0$, and suppose that $H_1(\mu_n)\to0$ so slowly that $$H_1(\mu_n)\ge1\Big/\sqrt{\ln\frac1{t_n}}$$ eventually (i.e., for all large enough $n$).

Then $\mu_n(j)\le t_n$ for $j\ge2$ and therefore $$\ln^2\frac1{\mu_n(j)}\ge\ln\frac1{t_n}\;\ln\frac1{\mu_n(j)},$$ whence eventually $$\begin{aligned} H_2(\mu_n)&\ge\sum_{j\ge2}\mu_n(j)\ln^2\frac1{\mu_n(j)} \\ &\ge\sum_{j\ge2}\mu_n(j)\ln\frac1{\mu_n(j)}\; \ln\frac1{t_n} \\ &=\Big(H_1(\mu_n)-(1-t_n)\ln\frac1{1-t_n}\Big)\ln\frac1{t_n} \\ &\ge \Big(1\Big/\sqrt{\ln\frac1{t_n}}(1-o(1))\Big)\ln\frac1{t_n}\to\infty. \end{aligned} $$

Thus, whenever 1' and 3' hold, 2' cannot hold.