Prove that for all $n\in\mathbb N^+ $ that
$2n \choose n $ $\ge 2^{2n-1}/ n^{1/2}$
I know that I am suppose to use induction but I am having trouble getting past what I have:
Proof:
Suppose that $p(n) $ : $2n \choose n $ $\ge 2^{2n-1}/ n^{1/2}$
Base case $p(1)$ is clearly true.
Induction: $p(n) \Rightarrow p(n+1)$
$2n+1 \choose n+1 $ = $ 2n \choose n+1$ + $2n \choose n$
By the inductive hypothesis we need to know prove that:
$2n+1 \choose n+1 $ $\ge 2^{2n-1}/ n^{1/2}$ + $2n \choose n$
Is this correct? How do I go about doing this from this point? Can someone provide me with the full proof?
$ P(n) \implies P(n+1)$ : $$ \binom {2n}{n} > \frac{2^{2n-1}}{\sqrt n}$$ $$\frac{2(2n+1)}{(n+1)} \binom{2n}{n}> \frac{2^{2n}(2n+1)}{(n+1)\sqrt n}$$ But $$ \frac{2^{2n}(2n+1)}{(n+1)\sqrt n} > \frac{2^{2n+1}}{\sqrt {(n+1)}}$$ Which is true as can be seen after some algebra and so one has: $$\binom {2n+2}{n+1} =\frac{2(2n+1)}{(n+1)} \binom{2n}{n}> \frac{2^{2n}(2n+1)}{(n+1)\sqrt n}>\frac{2^{2n+1}}{\sqrt {(n+1)}}$$ As wanted