Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$
What I've come up with is the following: 
$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$
$\underline{n=k:}\qquad $$$\frac {4^k}{k+1} \lt \frac{(2k)!}{(k!)^2}  $$
$\underline{n+1}:$ $$\frac {4^{k+1}}{k+2} \lt \frac{(2(k+1))!}{((k+1)!)^2} $$
 $$\frac{4^k \cdot 4}{k+2} \lt \frac {(2k+2)!}{(k!)^2\cdot(k+1)\cdot(k+1)}$$
$$\frac{4^k \cdot 4}{k+2} \lt  \frac{(k+1)\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot(2k+2)}{(k!)^\require{enclose}\enclose{updiagonalstrike}2\cdot(k+1)\cdot(k+1)}$$
$$\frac{4^k \cdot 4}{k+2} \lt  2 \cdot\frac{\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}{(k!)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}$$
$$\frac{4^k \cdot 4}{k+2} \lt  2 \cdot\frac{{(k+2)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)}{(k!)}$$
From here on I'm stuck. Can somebody help me please? 
 A: For $k+1$ it is equivalent to
$$
\frac{4^k}{k+1}<\frac{(2k+1)(k+2)(2k)!}{2(k+1)^2(k!)^2}$$
It is enough to prove that
$$\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$
Notice that the above fraction is monotonically decreasing and its limit at $k\to\infty$ is $1$, therefore it is always greater than $1$.
And also, you have a mistake, as Harry noticed, $(2k)!\neq2k!$.
Edit
I have no idea what is Jaroslaw talking about. There is step-by-step solution. Suppose it is true for $k$. Then it is true for $k+1$ if and only if
$$\frac{4^{k+1}}{k+2}<\frac{(2k+2)!}{\left((k+1)!\right)^2}\\\frac{4^k}{k+1}\cdot\frac{4(k+1)}{k+2}<\frac{(2k+2)(2k+1)(2k)!}{(k+1)^2(k!)^2}\\\frac{4^k}{k+1}\cdot\frac{4(k+1)}{k+2}<\frac{(2k)!}{(k!)^2}\cdot\frac{(2k+2)(2k+1)}{(k+1)^2}$$
It is enough to prove that:
$$\frac{4(k+1)}{k+2}<\frac{(2k+2)(2k+1)}{(k+1)^2}$$
It is equivalent to
$$\frac{(2k+2)(2k+1)}{(k+1)^2}>\frac{4(k+1)}{k+2}\\\frac{2(k+1)(2k+1)(k+2)}{4(k+1)(k+1)^2}>1\\\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$
A: Note that $$\frac{\frac{4^{n+1}}{n+2}}{\frac{4^n}{n+1}}=4\frac{n+1}{n+2}$$ and that $$\frac{\frac{(2(n+1))!}{(n+1)!^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)!}{(2n)!}\cdot\frac{n!^2}{(n+1)!^2}=\frac{(2n+2)(2n+1)}{(n+1)^2}=2\frac{2n+1}{n+1}. $$So, when is it true that$$4\frac{n+1}{n+2}\leqslant2\frac{2n+1}{n+1}?$$ Well, the previous inequaliy is equivalent to $2(n+1)^2\leqslant(2n+1)(n+2)=2n^2+5n+2$, which clearly holds always. So, since you proved that $\frac{4^2}{2+1}<\frac{(2\times2)!}{2!^2}$, the inequality that you want to prove is true whenever $n\geqslant2$.
A: As your first steps of mathematical induction are right, I'll just perform a proof of inductive thesis for $n=k+1$.
First notice, that for $k>1$
$$\frac{4(k+1)}{(k+2)} < \frac{(2k+2)(2k+1)}{(k+1)^2}$$
proof:
$$(k-1)k(k+1)^2(k+2)>0\\
\frac{k(k-1)}{(k+1)^2(k+2)}>0\\
\frac{4k^2-4k}{(k+1)^2(k+2)}>0\\
\frac{-4k^2+4k}{(k+1)^2(k+2)}<0\\
\frac{4(k^3+3k^2+3k+1)-(4k^3+16k^2+8k+4)}{(k+1)^2(k+2)}<0\\
\frac{4(k+1)^3}{(k+1)^2(k+2)}<\frac{4k^3+16k^2+8k+4}{(k+1)^2(k+2)}\\
\frac{4(k+1)^2}{(k+1)^2(k+2)}<\frac{(2k+2)(2k+1)(k+2)}{(k+1)^2(k+2)}\\
\frac{4(k+1)}{(k+2)}<\frac{(2k+2)(2k+1)}{(k+1)^2}$$
Thus, since $\frac{4(k+1)}{(k+2)}$ and $\frac{4^k}{k+1}$ are both positive, we have:
$$\frac{4^{k+1}}{k+2}=\frac{4(k+1)}{(k+2)}\frac{4^k}{k+1} < \frac{4(k+1)}{(k+2)}\frac{(2k)!}{(k!)^2}<\frac{(2k+2)(2k+1)}{(k+1)^2}\frac{(2k)!}{(k!)^2} = \frac{(2k+2)!}{(((k+1)!)^2}$$
