Evaluating the following limit

Question

Evaluate the following limit: lim_n→∞{(4ⁿ-C(2n+1,n))/(4ⁿ)} Where C(a,b) = a!/(a-b)!b!

Hint: Use asymptotics to bound the limit and then make use of the squeeze theorem.

Spoiler: The answer is 1.

Answer 1

lim_n→∞{(4ⁿ-C(2n+1,n))/(4ⁿ)}

=lim_n→∞{1-(C(2n+1,n))/(4ⁿ)}

=lim_n→∞{1-(((2n+1)/(n+1))C(2n,n))/(4ⁿ)}

=1- lim_n→∞{(((2n+1)/(n+1))C(2n,n))/(4ⁿ)}

By asymptotics we know lim_n→∞((2n+1)/(n+1))C(2n,n))/4ⁿ ≤ lim_n→∞ (4ⁿ√(2n+1)/(n+1))/4ⁿ

an necessarily, 0 ≤ lim_n→∞((2n+1)/(n+1))C(2n,n))/4ⁿ

Evaluating lim_n→∞ (4ⁿ√(2n+1)/(n+1))/4ⁿ yields,

lim_n→∞ (4ⁿ√(2n+1)/(n+1))/4ⁿ

⁼lim_n→∞√(2n+1)/(n+1) (note: the denominator will out grow the numerator)

= 0

this gives us,

0 ≤ lim_n→∞((2n+1)/(n+1))C(2n,n))/4ⁿ ≤ lim_n→∞ (4ⁿ√(2n+1)/(n+1))/4ⁿ

0 ≤ lim_n→∞((2n+1)/(n+1))C(2n,n))/4ⁿ ≤ 0

and by squeeze theorem, lim_n→∞((2n+1)/(n+1))C(2n,n))/4ⁿ=0

And finally we can see that,

1- lim_n→∞{(((2n+1)/(n+1))C(2n,n))/(4ⁿ)}

=1- 0

=0

Thus we have shown,

lim_n→∞{(4ⁿ-C(2n+1,n))/(4ⁿ)} = 1. Q.E.D