Why do these expressions tend to zero? 
*

*In a chapter on Markov chains, it is claimed that $\binom {2n}{n} p^n(1-p)^n$ (where $p$ is a probability) tends to $0$ as $n$ tends to $\infty$. But why is this so?

*It is also claimed that for an unrestricted simple random walk where the only possible steps are $+1$ or $-1$, all n-step transition probabilities tend to $0$ as $n$ tends to $\infty$. Again, why is this so?
 A: We will use the result

if $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=a $ and $|a|<1$, then $\lim_{n\to \infty}{a_n}=0$.

Let 
$$ a_n = \binom {2n}{n} p^n(1-p)^n \implies \frac{a_{n+1}}{a_n}= {\frac { 2p(1-p)\left( 2\,n+1 \right)  }{(n+1)}}. $$
Taking the limit of the last expression 
$$ \lim_{n\to \infty} \frac{a_{n+1}}{a_n}= 4 p(1-p).$$
Now, note that, $f(p)=4p(1-p)<1$ for $0<p<1$ and $p\neq \frac{1}{2}$. It is clear that $p=\frac{1}{2}$ is a special case since $\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = 1$. 
A: *

*We can show that ${2n \choose n}p^n(1-p)^n \rightarrow 0$ as $n \rightarrow \infty$ by the ratio test. Let $a_n = {2n \choose n}p^n(1-p)^n$, then
$$
\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n} = \lim_{n \rightarrow \infty} p(1-p) \frac{(2n+2)(2n+1)}  {(n+1)^2} = 4p(1-p)
$$
Since $p$ is a probability, $4p(1-p) \leqslant 1$, if $p \neq \frac{1}{2}$, we're done and the ratio test is successful. If $p = \frac{1}{2}$, the ratio test is inconclusive, and this actually gets very tricky indeed. I'm not sure how to resolve that.

*This second part is, naturally, closely related to the first, but we must also assume that $p \notin \{0,1\}$. Suppose we take $t$ steps from our starting point $x_0$. Let $X_t$ be the random variable representing the state at time $t$. If we make $i$ upward steps, then our final resting place is $x_0 + i - (t-i) = x_0 + 2i - t$. The number of upward steps we make is distributed binomially, so we can show that
$$
\mathbb{P}(X_t = x_0 - t + 2i) = 
\begin{cases}
{t \choose i}p^i(1-p)^{t-i} & \mbox{for $i \in \{0,1,...,t\}$} \\
0 & \mbox{otherwise}
\end{cases}
$$
We now adapt the argument from part 1. Fix an $i$ and let $t$ tend to infinity. We can assume that $i<t$. Let $a_t = {t \choose i}p^i(1-p)^{t-i}$, then
$$
\lim_{t \rightarrow \infty}\frac{a_{t+1}}{a_t} = \lim_{t \rightarrow \infty} (1-p) \frac{t+1}  {t+1-i} = (1-p)
$$
Since $0< p < 1$, $|(1-p)| < 1$, and these probabilities again tend to 0.
In the edge cases where either $p=0$ or $p=1$, the random walk just runs off in some direction or other, and we know its exact location after $t$ steps. The probabilities become Kronecker deltas, which trivially tend to 0.
A: Since $p(1-p)\le \frac14$ it suffices to show ${2n\choose n}4^{-n}\to 0$. This is the probability of getting exactly $n$ tails with a fair coin in $2n$ tosses. The probability of $n+k$ tails is ${2n\choose n+k}4^{-n}={2n\choose n}4^{-n}\cdot \frac{n(n-1)\cdots(n-k+1)}{(n+1)(n+2)\cdots (n+k)}$.
For fixed $k$ the fraction on the right tends to $1$ as $n\to\infty$, hence for sufficiently big $n$, we have 
If $n$ is big enough, ${2n\choose n+k}4^{-n}>\frac12 {2n\choose n}4^{-n}$. From this we find by adding probabilities that for any $k$ that $(1+\frac k2){2n\choose n}4^{-n}<1$ for $n$ big enough.
A: Using the asymptotic expansion (either via Stirling or Catalan number $C_n$): $$\binom {2n}{n} = 4^n \left(\frac{1}{\sqrt{n \pi}}+ O\left(n^{-\frac{3}{2}}\right)\right)$$
you get with $p(1-p) \le \frac{1}{4}$ for $0\le p\le 1$ an asymptotic estimate for your expression
$$\binom {2n}{n} p^n(1-p)^n = \binom {2n}{n} \left(p(1-p)\right)^n  \le \frac{1}{\sqrt{n \pi}}$$
