How can the estimate $ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1} $be made?

Question

Let's define the average value of binomial coefficients by $$ A (n) = \frac {1}{n+1} \sum_{i=0}^n \binom{n}{i} $$ Clearly, $$A= \frac {2^n}{n+1} \qquad .$$ Now, this gives us the trivial estimate for $$ \binom{n}{\frac {n}{2}} $$ as $$ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1} $$ I know that the largest possible value of $\binom{n}{k}$ is at $k= \frac {n}{2}$

The estimate holds just because this largest value is greater than the average value of binomial coefficients? Is it enough to say that this estimation holds since the maximum number in a set is always at least the average value?

What are you asking? Are you looking for a better estimate of $\binom{n}{\lfloor n/2\rfloor}$? — Mike Earnest, Jul 27 '17 at 18:24
It is the largest term of $2^n=(1+1)^n=\sum_k\binom{n}{k}$. If the maximum term is smaller than $\frac{2^n}{n+1}$ then the whole sum is smaller than $2^n$. — Hellen, Jul 27 '17 at 18:26
The maximum number in a set is always at least the average value. — symplectomorphic, Jul 27 '17 at 18:26
Hint: $\binom{n}{\lfloor n/2\rfloor}\geq \binom{n}{i}$ for all $i$. — Thomas Andrews, Jul 27 '17 at 18:27
Is it enough to say that this estimation holds since the maximum number in a set is always at least the average value.? @symplectomorphic — Leyla Alkan, Jul 27 '17 at 18:37

score 3 · Accepted Answer · answered Jul 27 '17 at 19:02

3

Sure that's valid! It's the same as saying $$ \frac{2^n}{n+1} = \frac1{n+1}\sum_{i=0}^n\binom{n}i\leq \frac1{n+1}\sum_{i=0}^n\binom{n}{n/2} = \frac{n+1}{n+1}\binom{n}{n/2} = \binom{n}{n/2}. $$

answered Jul 27 '17 at 19:02

Tai

2,235

It's also valid $\binom{n}{\frac{n}{2}}<\frac{2^{n+1}}{n}$ I found expanding at infinity this nice estimate, quite precise $\binom{n}{\frac{n}{2}} \approx 2^n \sqrt{\frac{2}{\pi n}}$ – Raffaele Jul 27 '17 at 19:13
@Raffaele There are even tighter bounds, as seen from here. – Simply Beautiful Art Jul 27 '17 at 19:55

Michael Rozenberg · Answer 2 · 2017-07-27T18:46:32.303

2

Just induction!

Firstly, we'll replace $n$ at $2n$.

Thus, we need to prove that $\binom{2n}{n}\geq\frac{2^{2n}}{2n+1}.$

For $n=1$ it's obvious.

Let $$\binom{2n}{n}\geq\frac{2^{2n}}{2n+1}.$$ Hence,$$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}\geq\frac{2^{2n}}{2n+1}\cdot\frac{2(2n+1)}{n+1}=$$ $$=\frac{2^{2n+2}}{2n+3}\cdot\frac{2n+3}{2(n+1)}>\frac{2^{2n+2}}{2n+3}$$ and we are done!

edited Jul 27 '17 at 18:46

answered Jul 27 '17 at 18:41

Michael Rozenberg

194,933

How can the estimate $ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1} $be made?

2 Answers2