# Finite sum and limit invoving binomial coefficient

I have found some interesting formulas involving binomial coefficients with the help of Mathematica. But I have no idea how it did. Could anyone help me?

Here they are:

$$\sum_{k=0}^{m-1} 2^{-2k} \binom{2k}{k} = m 2^{1-2m} \binom{2m}{m}\tag{1}$$

$$\lim_{m \rightarrow \infty} \sqrt{m} 2^{1-2m} \binom{2m}{m} = \frac{2}{\sqrt{\pi}}\tag 2$$

Thank you.

Update:

Stirling's approximation shows $$\binom{2m}{m} \sim \frac{4^m}{\sqrt{\pi m}}$$ as $m \rightarrow \infty$. Substitution of it solves the problem 2.

• for the second one, try stirling's formula! – Ant Oct 7 '15 at 8:24
• Instead of updating your question, it is better to post a (partial) answer. There is absolutely no problem with answering your own question, and this also helps to reduce the number of unanswered questions. – zhoraster Oct 7 '15 at 8:57

## 1 Answer

(1) can be rewritten as $$\sum_{k=0}^{m-1} {2k \choose k}2^{2(m-k)} = 2m {2m \choose m}.$$ Thinking generatingfunctionally, the lhs is a coefficient of $x^m$ in the product of $$A(x) = \sum_{k=0}^\infty {2k \choose k} x^k = \frac{1}{\sqrt{1-4x}}$$ and $$B(x) = \sum_{k=1}^\infty 2^{2k} x^k = \frac{4x}{1-4x}.$$ The rhs is a coefficient of $x^m$ in $$\sum_{k=0}^\infty 2k{2k \choose k} x^k = \frac{4x}{(1-4x)^{3/2}}.$$

Hence the equality.

I was trying to come up with some combinatorial argument, but failed.