Prove that for n ≥ 2 and k ≤ n, ${n \choose k} ≤ \frac{n^k}{2^{k−1}}$

Question

Is this bound ever tight, meaning does equality ever hold for some n, k?

I am not sure how to proceed with this inequality, any direction will help greatly.

Answer 1

A simple set of inequalities for $n!$: $n! \ge 2^{n-1}$; for $n \ge 3$, $n! \ge 2\cdot 3^{n-2}$; for $n \ge m$, $n! \ge m!\cdot (m+1)^{n-m}$.

Proof: $n! =\prod_{k=1}^n k =(\prod_{k=1}^m k)( \prod_{k=m+1}^n k) \ge m! (m+1)^{n-m} $.

For $m=1$ this is $n! \ge 2^{n-1}$.

For $m=2$ this is $n! \ge 2\cdot 3^{n-2}$.

We have

$\binom{n}{k} =\dfrac{n!}{k!(n-k)!} =\dfrac{\prod_{j=0}^{k-1}(n-j)}{k!} \le\dfrac{n^k}{m!(m+1)^{k-m}} $ for $k > m$.

For $m=1$, this is your inequality.

If we use $n! > (n/e)^n$ (proved here a number of times), then

$\binom{n}{k} =\dfrac{\prod_{j=0}^{k-1}(n-j)}{k!} \le\dfrac{n^k}{(k/e)^k} =(en/k)^k $ and this is stronger for $k \ge 2e$ or $k \ge 6$.