Mathematical Induction Inequality Proof with Factorials

Worked Example

Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2 \).

Step 1:  Show it is true for \( n =2 \).
\( \begin{aligned} \require{color}
\text{LHS } &= (2 \times 2)! = 16 \\
\text{RHS } &= 2^2 \times (2!) = 8 \\
\text{LHS } &> { RHS} \\
\end{aligned} \)
\( \therefore \text{It is true for } n =2 \)
Step 2:  Assume it is true for \(n =k\), that is \( (2k)! > 2^k (k!)^2. \)
Step 3:  Show it is true for \(n =k+1\), that is \( (2k+2)! > 2^{k+1} \big[(k+1)!\big]^2. \)
\( \begin{aligned} \require{color}
\text{LHS } &= (2k+2)! \\
&= (2k+2)(2k+1)(2k)! \\
&= 2(k+1)(2k+1)(2k)! \\
&> 2(k+1)(2k+1)2^k (k!)^2 &\color{red} \text{by assumption} \\
&> (k+1)(2k+1)2^{k+1} (k!)^2 \\
&> (k+1)(k+1)2^{k+1} (k!)^2 &\color{red} 2k+1>k+1 \\
&= 2^{k+1} \big[(k+1)k!\big]^2 \\
&= 2^{k+1} \big[(k+1)!\big]^2 \\
&= \text{RHS} \\
\end{aligned} \\ \)
\( \therefore (2k+2)! > 2^{k+1} \big[(k+1)!\big]^2 \text{ for } \ge 2.\)

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