For a random variable $X$ s.t. $X$ has a Poisson distribution:
$$ P(k \text{ events in interval}) = \frac{\lambda^k e^{-\lambda}}{k!} $$
The following graph seems to indicate that the maximum probability value is $k = \lambda$ (and perhaps also $\lambda-1$).
Question: Formally, why is this the case?
I have been able to do some algebraic manipulation:
$$ f_X(x) = \frac{e^{-\lambda}\lambda^x}{x!} = \frac{\lambda^x}{x! \sum_{n=0}^{\infty}\frac{\lambda^n}{n!}} $$
so that
$$ f_X(x) = \frac{\lambda^x}{\lambda^x + x! \left( \sum_{n=0}^{x-1} \frac{\lambda^n}{n!} + \sum_{m=x+1}^{\infty} \frac{\lambda^m}{m!} \right)} $$
But it is still not clear to me why the maximum is at $x = \lambda$ and $x = \lambda-1$.
For $k\ge 0$, we calculate the ratio $$\frac{\Pr(X=k+1)}{\Pr(X=k)}.\tag{1}$$ This is $$\frac{e^{-\lambda}\frac{\lambda^{k+1}}{(k+1)!}}{e^{-\lambda}\frac{\lambda^k}{k!}},$$ which simplifies to $$\frac{\lambda}{k+1}.$$ Thus if $\lambda \lt 1$, then the ratio (1) is $\lt 1$ at $k=0$, and even smaller afterwards. Thus $\Pr(X=k)$ is steadily decreasing, and therefore our probability attains a maximum at $k=0$.
If $\lambda$ is a positive integer, then the ratio (1) is greater than $1$ if $k+1\lt \lambda$, is equal to $1$ if $k+1=\lambda$, and is less than $1$ if $k+1\gt \lambda$. Thus our probability attains a maximum at $k=\lambda-1$ and also at $k=\lambda$.
Finally, suppose $\lambda\gt 1$ and $\lambda$ is not an integer. Then the ratio (1) is greater than $1$ if $k-1\lt \lambda$, and is less than $1$ if $k\gt \lambda$, so our probability attains a maximum at $k=\lfloor \lambda\rfloor$.