Ratio of the first squared and the second moment 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:
$$
G(t)=\frac{6}{\pi^2}\text{PolyLog}(2,t).
$$
 A: 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 $C=0$.
