Let $\lambda_n=1/n$ for $n=1,2,\ldots$.Let $X_n \sim \text{Poisson}(\lambda_n)$.
(a) Show that $X_n \xrightarrow{\text{P}} 0$.
(b)Let $Y_n =nX_n$. Show that $Y_n \xrightarrow{\text{P}} 0$.
For part (a) I've used Markov's inequality $$\mathbb{P}(\left | X_n \right |>\varepsilon )\leq \frac{\mathbb{E}[X_n]}{\varepsilon }=\frac{\lambda _n}{\varepsilon }= \frac{1}{n\varepsilon }\xrightarrow[n\rightarrow \infty ]{ } 0$$
But for part (b) nothing seems to work. Markov's inequality leads to $$\mathbb{P}(\left | Y_n \right |>\varepsilon )\leq \frac{\mathbb{E}[nX_n]}{\varepsilon }=\frac{1}{\varepsilon } $$ Neither does Chernoff's method $$ \mathbb{P}\left (X_n>\frac{\varepsilon}{n} \right ) =\mathbb{P}\left (e^{tX_n}>e^{t\varepsilon/n} \right ) \leq \frac{\mathbb{E}[e^{tX_n}]}{e^{t\varepsilon/n}}= \frac{e^{(e^t-1)/n}}{e^{t\varepsilon/n}}= \xrightarrow[n\rightarrow \infty ]{ } 1 $$ And if I haven't done any mistakes, $Y_n$ does not converge in $L^p$ to $0$.
At this point I'm stuck. I'd appreciate any advice.
You can do it using only elementary facts about $(X_n)_n$: fixing any $\varepsilon >0$, for $n\geq 1$ $$\begin{align} \mathbb{P}\{|Y_n| > \varepsilon\} &=\mathbb{P}\{Y_n > \varepsilon\}\\ &\leq \mathbb{P}\{Y_n > 0\} \\ &= \mathbb{P}\{X_n > 1/n\} \\ &= \mathbb{P}\{X_n \geq 1\} \tag{$X_n$ is integer-valued}\\ &= 1-\mathbb{P}\{X_n =0\}\\ &= 1-e^{-\lambda_n}\\ &= 1-e^{-1/n}\\ &\xrightarrow[n\to\infty]{} 0 \end{align}$$
On the other hand, you are right, I don't see how to use the MGF approach to easily make this work.