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.



Existence of sequence of distributions
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.


