prove by induction that ${ 2n \choose n}\ge \frac{4^n}{2n+1}$

Question

I need to show that by induction that ${ 2n \choose n}\ge \frac{4^n}{2n+1}$. for the base case it's trivial but for n=k+1 and using the induction hypothesis I got $\frac{(2k+2)(2k+1)}{(k+1)}\cdot {2k \choose k}$. now I got stuck and can't continue

Answer 1

See that $$\frac{(2k + 2)(2k + 1)}{(k + 1)^2}\cdot{{2k}\choose{k}} \ge \frac{4^{k+1}}{2k + 3} = \frac{4\cdot4^{k}}{\frac{2k+3}{2k + 1}(2k + 1)}$$

The part in red is already true:

$$\frac{2(2k + 1)}{(k + 1)}\cdot\color{red}{{{2k}\choose{k}}} \ge \frac{4\cdot\color{red}{4^{k}}}{\frac{2k+3}{2k + 1}\color{red}{(2k + 1)}}$$

Hence, we just have to prove that $$\frac{2(2k + 1)}{(k + 1)} > \frac{4}{\frac{2k+3}{2k + 1}} \text{ for } \color{blue}{k \ge 1}$$

Simplifying, we obtain

$$\frac{2(2k + 1)}{k + 1} > \frac{4(2k + 1)}{2k + 3}$$

$$\frac{2}{k + 1} > 0$$

$$k > -1$$

This tells us the inequality is true for $k > -1$. Hence, the inequality is also true for $k \ge 1$.

Ideally, $$\frac{2(2k + 1)}{k + 1} > 0 \text{ which gives } k > -\frac{1}{2}$$ Even though $k = -\frac{1}{2}$ were ommitted due to division, they are our not in our desired domain and does not affect our answer, so this restriction can be ignored.

Answer 2

We need to prove the following result:

For every non-negative integer $m$ and for any integer $n$ such that $0 \leq n \leq 2m$, we have $$ { 2m \choose n} \geq \frac{4^n}{2n+1}. \tag{0} $$

Proof:

Let us first put $m := 0$. Then we have $0 \leq n \leq 2(0) = 0$, that is, $n = 0$, and then $$ \begin{align} { 2m \choose n } &= { 0 \choose 0} \\ &= \frac{ 0!}{0! (0-0)!} \\ &= 1 \\ &= \frac{ 4^0}{ 2(0) + 1}. \end{align} $$ So that (0) above holds for $m = 0$.

Next suppose that $m$ is some fixed non-negative integer such that (0) above holds for all integers $n$ such that $0 \leq n \leq 2m$.

Now we have the following three cases: Case 1. $0 \leq n \leq 2m$. Case 2. $n = 2m+1$. Case 3. $n = 2m+2$.

Case 1. When $n$ is an integer between $0$ and $2m$ inclusive, then we have $$ \begin{align} & \qquad { 2(m+1) \choose n } \\ &= { 2m+2 \choose n } \\ &= \frac{ (2m+2)! }{ n! (2m+2 - n)!} \\ &= \frac{ (2m+2)(2m+1) (2m)!}{ n! (2m+2 -n ) (2m+1-n) (2m-n)!} \\ &\qquad \qquad \mbox{[ Note that $0 \leq n < 2m+1$ ]} \\ &= \frac{(2m+2)(2m+1)}{ (2m+2-n)(2m+1-n) } \frac{ (2m)!}{ n! (2m-n)!} \\ &= \frac{(2m+2)(2m+1)}{ (2m+2-n)(2m+1-n) } { 2m \choose n } \\ &\geq \frac{(2m+2)(2m+1)}{ (2m+2-n)(2m+1-n) } \frac{4^n }{ 2n+1 } \\ &\geq \frac{4^n }{ 2n+1 } \\ &\qquad \mbox{[ as $0 \leq n < 2m+1$, so we have $\frac{(2m+2)(2m+1)}{ (2m+2-n)(2m+1-n) } > 1$ ]}. \end{align} $$

Case 2. For $n = 2m+1$, we have $$ \begin{align} { 2(m+1) \choose n } &= { 2m+2 \choose 2m+1 } \\ &= \frac{ (2m+2)!}{ (2m+1)! \big( (2m + 2) - (2m+1) \big)!} \\ &= \frac{ (2m+2)(2m+1)!}{ (2m+1)! 1!} \\ &= 2m+2 \\ &\not\geq \frac{ 4^{2m} }{ 2(2m) + 1} = \frac{ (16)^m }{ 4m+1} \end{align} $$ unless $m = 0$ or $m=1$. So our induction argument breaks down.

Counter-Example:

Let $m = 2$ and $n = 3$. Then we have $$ { 2m \choose n } = { 4 \choose 3 } = 4, $$ but $$ \frac{ 4^n }{2n+1} = \frac{ 4^3}{2(3) + 1} = \frac{64}{7} > 4. $$

Note that we don't even need to consider Case 3 when $n = 2m+2$.

Answer 3

Assume $\cdot{{2k}\choose{k}} \ge \frac{4^{k}}{2k + 1}$ Then, $$ \cdot{{2(k+1)}\choose{k+1}} = \frac{(2k + 1)(2k + 2)}{(k + 1)^2}\cdot{{2k}\choose{k}} \ge \frac{4^{k}}{2k + 1} \frac{(2k + 1)(2k + 2)}{(k + 1)^2}= \frac{4^{k}.2}{k+1} = \frac{4^{k+1}}{2k+2} > \frac{4^{k+1}}{2k+3} = \frac{4^{k+1}}{2(k+1)+1} $$

This proves the inequality for n=k+1 given inequality is true for n=k