What's the answer to this limit question? Can anyone find the limit to this one?
$\lim_{n \rightarrow \infty} (\sum_{i=(n+1)/2}^n {n \choose i} \times 0.51^i \times 0.49^{n-i})$
When I plot it, it seems to me to approach 1, which makes me feel that a limit should exist and it should be 1. But I can't solve this mathematically.
Here is the same question in WolframAlpha, in case it helps.
 A: Consider tossing an unfair coin $2n-1$ times, where we assign the value $+1$ to head, which has probability $p=0.51$ and $-1$ to tails, with probability $0.49$. So we have a sequence of IID variables $X_1, \dots,X_{2n-1}$, and we denote the $S_n = \sum_{i=1}^{2n-1} X_i$. Furthermore, we have $$\mu = \mathbb{E} X_1 = 0.51 -0.49=0.02 >0 $$
Then we have $$ \mathbb{P}(S_n > 0)= \sum_{i=n}^{2n-1} \binom{2n-1}{i} p^i (1-p)^{2n-1-i}.$$
So we have 
\begin{eqnarray}
\lim_{n \to \infty} \sum_{i=n}^{2n-1} \binom{2n-1}{i} p^i (1-p)^{2n-1-i} &=& \lim_{n \to \infty } \mathbb{P} ( S_n > 0) \\
&=& \lim_{n \to \infty} \mathbb{P} \left( \frac{S_n}{2n-1} > 0\right)\\
&=& \lim_{n \to \infty } \mathbb{P} \left( \frac{1}{2n-1} \sum_{i=1}^{2n-1} X_i > 0 \right)
\end{eqnarray}
Now we can use the weak Law of Large Numbers and obtain
\begin{eqnarray} \lim_{n \to \infty} \mathbb{P} \left( \left|\frac{1}{2n-1} \sum_{i=1}^{2n-1} X_i  - \mu \right| > \epsilon \right) = 0
\end{eqnarray}
for all $\epsilon > 0$.
So if we choose $\epsilon = \frac{\mu}{2}>0$, we have $$\left\{ \frac{S_n}{2n-1} \in \left( \frac{ \mu }{2} , \frac{ 3 \mu}{2} \right) \right\} \subset \left\{ \frac{ S_n}{2n-1} > 0 \right\}.$$
And therefore we find
\begin{eqnarray}
1 &\ge& \lim_{n \to \infty } \mathbb{P} \left( \frac{ S_n}{2n-1} >0 \right) \\ &\ge & \lim_{n \to \infty} \mathbb{P} \left( \frac{S_n}{2n-1 } \in \left( \frac{\mu}{2} , \frac{ 3 \mu }{2} \right) \right) \\
&=& \lim_{n \to \infty } 1 - \mathbb{P} \left( \frac{S_n}{2n-1 } \not\in \left( \frac{\mu}{2} , \frac{ 3 \mu }{2} \right) \right) \\
&=& 1 - \lim_{n \to \infty} \mathbb{P} \left( \left| \frac{ S_n}{2n-1} - \mu \right| > \frac{\mu}{2} \right) \\
&=& 1 -0 = 1
\end{eqnarray}
So combining everything we see $$\lim_{n\to \infty} \sum_{i=n}^{2n-1} \binom{2n-1}{i} 0.51^p \times 0.49^{2n-1-i}=1$$
as desired.
