Is $\binom{n}{1}(2n - 1)!2^1 - \binom{n}{2}(2n - 2)!2^2 \cdots + (-1)^{n + 1}\binom{n}{n}n!2^n > \frac{(2n)!}{2}$ true?

Question

$$\binom{n}{1}(2n - 1)!2^1 - \binom{n}{2}(2n - 2)!2^2 + \binom{n}{3}(2n - 3)!2^3 - \cdots + (-1)^{n + 1}\binom{n}{n}n!2^n > \frac{(2n)!}{2}$$

Hello,

I want to prove/disprove the above inequality.

For $n = 1, 2, 3$, this works, so my guess is that it is correct.

I have tried proving by induction but it seems complicated. Maybe, the LHS is binomial expansion, I tried to bring the binomial coefficient and the factorial into one binomial coefficient, but I can't seem to do that. Other than that, I don't have much idea.

Is this inequality true? If yes, how can we prove it?

Thanks

Answer 1

The given inequality can be rewritten as $$\sum_{k=1}^{n} (-1)^{k+1} \binom{n}{k} \frac{(2n-k)!}{(2n)!} 2^k > \frac{1}{2}$$ Setting $a_k = \binom{n}{k} \frac{(2n-k)!}{(2n)!} 2^k$, compute that $$\frac{a_{k+1}}{a_k} = \frac{n-k}{2n-k} \frac{2}{k+1} < \frac{2}{3}$$ for $k\ge 2$. Thus, the sequence $\{a_k\}_{k\ge 2}$ is decreasing. We conclude that $$\sum_{k=1}^{n} (-1)^{k+1} a_k = a_1 + \sum_{k=2}^{n} (-1)^{k+1} a_k \ge a_1 - a_2 = 1 - \frac{n-1}{2n-1} > \frac{1}{2}$$ for all $n$.