Let $$P(X = k) = \frac{4}{k(k+1)(k+2)}$$ for $k ≥ 1$. I need to show that the generating function $φ_X(u) = E(u^X)$ satisfies $φ_X(1)=1,φ′_X(1_−)=2<∞$ , but $ φ′′ (1_−)=∞$.
By the way, in this case$|φ′ (u)|≤2$ uniformly in $|u|<1$ while still $φ_X(u) = ∞$ for all $|u| > 1$. If something is unclear, I am more than happy to add information and I am thankful for your help.
We compute the generating function:
\begin{align} \varphi_X(s) &= \sum _{k=1}^{\infty } \frac{4 s^k}{k (k+1) (k+2)}\\ &= \frac{3 s^2-2 s^2 \log (1-s)-2 s+4 s \log (1-s)-2 \log (1-s)}{s^2}. \end{align} From here we have $$ \lim_{s\uparrow1}\varphi_X(s) = 1 $$ and $$ \frac{\mathsf d}{\mathsf d s}\varphi_X(s) = \frac{-2 (s-2) s-4 (s-1) \log (1-s)}{s^3} $$ so $$ \lim_{s\uparrow1}\varphi_X'(s) = 2, $$ and $$ \frac{\mathsf d^2}{\mathsf d^2s}\varphi_X(s) = \frac{2 ((s-6) s+(4 s-6) \log (1-s))}{s^4} $$ so $$ \lim_{s\uparrow1}\varphi_x''(s) = \infty. $$ We can verify this as $1-\lim_{s\uparrow1}\varphi_X(s) = \mathbb P(X=\infty)=0$, $$ \mathbb E[X] = \sum_{k=1}^\infty \frac 4{(k+1)(k+2)} = 2, $$ and $$ \mathbb E[X^2] = \sum_{k=1}^\infty \frac {4k}{(k+1)(k+2)} = +\infty. $$