Let $X$ be a Poisson random variable with parameter $\lambda >0$.
Find $\sup_{k≥0}\Bbb P (X = k)$ and show that it goes to $0$ as $\lambda \to \infty$.
$\sup_{k≥0}\Bbb P (X = k) = \frac {\lambda ^k e^{-\lambda}} {k!}=\frac {\lambda ^k } {k!e^{\lambda}}$, which goes to $0$ as $λ \to \infty$ using lhopital rule, I believe.
Or I can used definition since the graph of a Poisson distribution with parameter finite a's probability gets to $0$ as the RV value approaches a. Does "sup" change any of this?
Note that $\frac {\lambda ^{k} e^{-\lambda}} {k!} <\frac {\lambda ^{k+1} e^{-\lambda}} {(k+1)!}$ if $k+1<\lambda$ and the reverse inequality holds if $k+1>\lambda$. It follows that the supremum of $\frac {\lambda ^{k} e^{-\lambda}} {k!}$ is attained at $k=[\lambda -1]$ where $[.]$ is the floor function. Now take $\lambda$ between $n$ and $n+1$ and show that $\frac {\lambda ^{(n-1)} e^{-\lambda}} {(n-1)!} \to 0$ uniformly for $\lambda$ in this range. This is easy from Stirling's approximation.