Let $G(t)$ be a probability generating function of some integer and non-negative r. v. $X$. Suppose that
$$\lim\_{t\to1}G'(t)=+\infty.$$
That is
$$
\mathbb{E}X=+\infty.
$$


Can you show that
$$
\lim\_{t\to1}\frac{(G'(t))^2}{G''(t)}=0\,?
$$


There is an example satisfying the conjecture:
$$
G(t)=\frac{6}{\pi^2}\text{PolyLog}(2,t).
$$



Ratio of the first squared and the second moment
This is correct.


Denote $G(t)=\sum\_{k=0}^\infty p\_k t^k$, where $p\_i\geqslant 0$ and $\sum p\_i=1$. Then we are given that $\sum kp\_k=\infty$ and should prove that
$$
\lim\_{t\to 1-0} \frac{(\sum\_{k=1}^\infty kp\_k t^{k-1})^2}{\sum\_{k=2}^\infty k(k-1)p\_kt^{k-2}}=0.
$$
Since both numerator and denominator tend to infinity when $t$ goes to $1-0$, we have for arbitrary positive integer $N>1$:
$$C:=\limsup\_{t\to 1-0} \frac{(\sum\_{k=1}^\infty kp\_k t^{k-1})^2}{\sum\_{k=2}^\infty k(k-1)p\_kt^{k-2}}=\limsup\_{t\to 1-0} \frac{(\sum\_{k=N}^\infty kp\_k t^{k-1})^2}{\sum\_{k=N}^\infty k(k-1)p\_kt^k}\\
=\left(\sum\_{k=N}^\infty p\_k\right)\cdot \limsup\_{t\to 1-0} \frac{(\sum\_{k=N}^\infty kp\_k t^{k-1})^2}{(\sum\_{k=N}^\infty k(k-1)p\_kt^{k-2})(\sum\_{k=N}^\infty p\_k)}\leqslant 2 \sum\_{k=N}^\infty p\_k,
$$
since by Cauchy–Bunyakovsky–Schwarz we have
$$
\left(\sum\_{k=N}^\infty k(k-1)p\_kt^{k-2}\right)\left(\sum\_{k=N}^\infty p\_k\right)\geqslant \left(\sum\_{k=N}^\infty \sqrt{k(k-1)}p\_k t^{k-1}\right)^2
\geqslant \frac12
\left(\sum\_{k=N}^\infty kp\_k t^{k-1}\right)^2.
$$
It remains to observe that $\sum\_{k=N}^\infty p\_k$ may be arbitrarily small, thus $C=0$.


