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.


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.

  • $\begingroup$ for the second one, try stirling's formula! $\endgroup$ – Ant Oct 7 '15 at 8:24
  • $\begingroup$ 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. $\endgroup$ – zhoraster Oct 7 '15 at 8:57

(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.

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