I'm looking at a question where I have to prove that:
$$(1-z)^{-1/2}=\sum_{k=0}^\infty\binom{2k}k\frac{z^k}{2^{2k}}$$
The hint is to use the generalised Binomial Theorem. I'm having trouble getting the expected answer.
https://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem has: $$\frac{1}{(1-z)^s}=\sum_{k=0}^\infty\binom{s+k-1}{k}z^k$$
So as I see it, the series should be:
$$(1-z)^{-1/2}=\sum_{k=0}^\infty\binom{k-1/2}kz^k$$
$$=\sum_{k=0}^\infty\frac{1/2(1/2-1)(1/2-2)…(1/2-k+1)}{k!}z^k$$
$$=\sum_{k=0}^\infty\frac{(1-2)(1-4)…(1-2k+2)}{k!}\frac{z^k}{2^k}$$
$$=\sum_{k=0}^\infty\frac{(3)(5)(7)…(2k-3)}{k!}\frac{-(-z)^k}{2^k}$$
$$=\sum_{k=0}^\infty\frac{(2k-3)!}{k!(2)(4)…(2k-4)}\frac{-(-z)^k}{2^k}$$
$$=\sum_{k=0}^\infty\frac{(2k-3)!}{k!(k-2)!}\frac{-(-z)^k}{2^k2^{k-2}}$$
$$=\sum_{k=0}^\infty\frac{2k!}{k!(k-2)!}\frac{-(-z)^k}{2^k2^{k-2}(2k-2)(2k-1)2k}$$
$$=\sum_{k=0}^\infty\binom{2k}k\frac{-(-z)^kk(k-1)}{2^k2^{k-2}(2k-2)(2k-1)2k}$$
$$=\sum_{k=0}^\infty\binom{2k}k\frac{-(-z)^k}{2^{2k}(2k-1)}$$
So I've ended up with something close, but have an extra factor and some sign issues. Any clues as to where I've gone wrong?
Best Answer
By the generalized binomial theorem, $$ (1-z)^{-1/2}= \sum_{k=0}^\infty\binom{-1/2}{k}(-z)^k $$ so your task is to prove that $$ (-1)^k\binom{-1/2}{k}=\frac{1}{2^{2k}}\binom{2k}{k} $$ This is surely true for $k=0$. Next, by induction, $$ (-1)^{k+1}\binom{-1/2}{k+1}= -(-1)^k\binom{-1/2}{k}\frac{-1/2-k}{k+1}= \frac{1}{2^{2k}}\binom{2k}{k}\frac{1}{2}\frac{2k+1}{k+1} $$