Chung's Elementary probability theory page 225 
As a consequence, for any fixed number $l$, we have
$$
\tag{7.3.8}\lim_{n\rightarrow\infty}\sum^l_{k=-l}\binom{2n}{n+k}\frac{1}{2^{2n}}=0
$$
because each term in the sum has limit $0$ as $n\rightarrow\infty$ as just shown, and there is only a fixed number of terms. Now if we remember Pascal's triangle (3.3.5), the binomial coefficients $\binom{2n}{n+k}$, $-n\leq k\leq n$, assume their maximum value $\binom{2n}{n}$ for the middle term $k=0$ and decrease as $|k|$ increases. According to (7.3.8), the sum of a fixed number of terms centered around the middle term approaches zero; hence a fortiori the sum of any fixed number of terms will also approach zero, namely for any fixed $a$ and $b$ with $a<b$, we have
$$
\tag{*}\lim_{n\rightarrow\infty}\sum^b_{j=a}\binom{2n}{j}\frac{1}{2^{2n}}=0\text{.}
$$

Can someone please explain the last sentence? I know why (*) is true, but I don't understand the author's argument. I think that is nothing to do with the middle term.
 A: The sum of any number of fixed terms is AT MOST equal to the sum of the same amount of terms centered at the centre $\binom{2n}{n}$, as is immediate from Pascal's triangle. Say, I want to sum 5 consecutive terms from the 8th row in Pascal's triangle, which goes $1,8,28,56,70,56,28,8,1$. There are many possibilities, like:
$1+8+28+56+70$
$8+28+56+70+56$
etc...
But the one with the greatest sum is always the one with it's centre at $70=\binom{8}{4}$, so the maximal value for the sum of 5 terms is $28+56+70+56+28$. With this logic, we have
$$0\leq\sum^b_{j=a}\binom{2n}{j}\leq\sum^l_{k=-l}\binom{2n}{n+k}$$
if $b-a$ is even (so there are an odd number of terms) and $l$ is defined to be $(b-a)/2$. We can multiply both sides by $1/2^{2n}$ and take the limit to get
$$0\leq\lim_{n\to \infty}\sum^b_{j=a}\binom{2n}{j}\frac{1}{2^{2n}}\leq\lim_{n\to \infty}\sum^l_{k=-l}\binom{2n}{n+k}\frac{1}{2^{2n}}=0$$
$$\lim_{n\to \infty}\sum^b_{j=a}\binom{2n}{j}\frac{1}{2^{2n}}=0$$
If $b-a$ is odd, we can compare it to a larger sum where the difference IS even, which leads to the same result.
A: It means that even if the range $(a,a+1,...,b)$ does not include the middle term $n$, in other words $n\notin\{a,a+1,...,b\}$, the sum will still approach zero, since it is then less than the sum which includes the middle term.
Call the sum from $a$ to $b$ $\sum_a^b$ and call $c=\min\{a,b,2n-a, 2n-b\}$ then $\sum_a^b \leq \sum_{c}^{2n-c}$ since all the terms of $\sum_a^b$ are in $\sum_{c}^{2n-c}$, and we know from (7.3.8) that $\sum_{c}^{2n-c}$ approaches zero.
