Let $\xi_1, \xi_2, \ldots$ be iid. standard normal random variables. (Recall their moment generating function: $\mathbb{E}\left(\mathrm{e}^{\lambda \xi_i}\right)=\mathrm{e}^{\lambda^2 / 2}$.) Let $a, b \in \mathbb{R}$, $$ S_n=\sum_{k=1}^n \xi_k, \quad \text { and } \quad X_n=\mathrm{e}^{a S_n-b n} . $$
Show that $$ X_n \rightarrow 0 \text { a.s. } \Leftrightarrow b>0 $$ but for any $r \geq 1$ $$ X_n \rightarrow 0 \text { in } \mathcal{L}^r \Leftrightarrow r<\frac{2 b}{a^2} $$
here's my solution but I'm stuck on how I can justify removing the $n$ from the final inequality. Might be having a bit of brain fart but it's late.
Now let $r \geq 1$, then using Markov's inequality: $$\mathbb{P}\left(\left|X_n-0\right| \geq \varepsilon\right)=\mathbb{P}\left(\left|X_n\right|^r \geq \varepsilon^r\right) \leq \frac{\mathbb{E}\left|X_n\right|^r}{\varepsilon^r}=\frac{\left\|X_n-X\right\|_r^r}{\varepsilon^r} \rightarrow 0$$
Using the moment generating function provided, $$\frac{\mathbb{E}\left|X_n\right|^r}{\varepsilon^r}=\frac{e^{r(\frac{a^2r}{2}-bn)}}{\varepsilon^r} \to 0.$$
By the definition of convergence, there exists an $N$ s.t $\forall n \geq N,$ $$\frac{e^{r\left(\frac{a^2 r}{2}-b n\right)}}{\varepsilon^r} \leq \frac{e^{r\left(\frac{a^2 r}{2}-b N\right)}}{\varepsilon^r}$$ so we choose $N$ s.t. $$\frac{e^{r\left(\frac{a^2 r}{2}-b N\right)}}{\varepsilon^r} =1 = e^0$$ then $$\frac{a^2 r}{2}-b n < 0$$
The "brain fart" happens when computing $E(X_n^r)$; notice that: \begin{align} E(X_n^r) &= e^{-bnr} E(e^{arS_n}) \\ &= e^{-bnr} \prod_1^n E(e^{ar \xi_1}) \\ &= e^{-bnr} \prod_1^n e^{\frac{a^2r^2}{2}} \\ &= e^{-bnr} e^{\frac{a^2r^2}{2} \color{red} n} \\ &= e^{\color{red} nr\left( \frac{a^2r^2}{2} - b \right)} \end{align} The rest of your proof carries on the same and you'll arrive at the right conclusion.