Prove that the following sequences monotnically decrease and increase correspondingly. Since they are bounded, find the limit.

First, let us make use of the following inequality.

Lemma. The function $x^k \mathrm{e}^{-x}$ is increasing on $[0, k]$ and decreasing on $[k, \infty)$.

Proof. The derivative is given by $k(k-x)x^{k-1}e^{-x}$, which is nonnegative for $0 \leq x \leq k$ and nonpositive for $x \geq k$.

Let's now show that $f_+$ decreasing. To start, note that we may write

$$f_+(n) = P(\xi_n \geq n) = \sum_{k=n}^{\infty} \frac{n^k \mathrm{e}^{-n}}{k!}; \quad\quad\quad f_-(n) = P(\xi_n < n) = \sum_{k=0}^{n-1} \frac{n^k \mathrm{e}^{-n}}{k!}$$
To check that $f_+$ is monotonically decreasing, it suffices to check that $f_+(n+1) < f_+(n)$ for all $n$. To see this, note that

$$f_+(n+1) = \sum_{k=n+1}^{\infty} \frac{(n+1)^k \mathrm{e}^{-(n+1)}}{k!} < \sum_{k=n+1}^{\infty} \frac{n^k \mathrm{e}^{-n}}{k!} < \sum_{k=n}^{\infty} \frac{n^k \mathrm{e}^{-n}}{k!} = f_+(n),$$

by the Lemma. Similarly, to show that $f_-$ is montonically increasing, it suffices to check that $f_-(n+1) > f_-(n)$ for all $n$. To see this, note that

$$f_-(n+1) = \sum_{k=0}^{n} \frac{(n+1)^k \mathrm{e}^{-(n+1)}}{k!} > \sum_{k=0}^{n} \frac{n^k \mathrm{e}^{-n}}{k!} > \sum_{k=0}^{n-1} \frac{n^k \mathrm{e}^{-n}}{k!} = f_-(n),$$
by the Lemma.

Now, to calculate the limit of both of these functions, note that the median $\nu$ of $\xi_n$ satisfies $|\nu - n| < 1$, meaning that the cumulative probabillity $f_+(n)$ is as close as possible to $1/2$, or in other words, $|f_+(n) - 1/2| < \frac{n^n\mathrm{e}^{-n}}{n!}$ (this is the difference in the cumulative provbability from $n$ to $n+1$). Taking the limit yields $\lim f_+ = 1/2$. Since $f_+ + f_- = 1$, we must also have $\lim f_- = 1/2$.