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Multiplicative Identity

2 n ! 22 nn !...

Question

$\frac{(2 n)!}{2^{2 n} (n!)^{2}} \leq \frac{1}{\sqrt{3 n + 1}}$ for all n ∈ N

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Solution

Let P(n) be the given statement.
Thus, we have:

$P (n) : \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \leq \frac{1}{\sqrt{3 n + 1}} Step 1 : P (1) : \frac{2!}{2^{2} . 1} = \frac{1}{2} \leq \frac{1}{\sqrt{3 + 1}} Thus, P (1) is true . Step 2 : Let P (m) be true . Thus, we have : \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} \leq \frac{1}{\sqrt{3 m + 1}} We need to prove that P (m + 1) is true .$

Now,

$P (m + 1) : \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} = \frac{(2 m + 2) (2 m + 1) (2 m)!}{2^{2 m} . 2^{2} {(m + 1)}^{2} {(m!)}^{2}} \Rightarrow \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} \leq \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} \times \frac{(2 m + 2) (2 m + 1)}{2^{2} {(m + 1)}^{2}} \Rightarrow \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} \leq \frac{2 m + 1}{2 (m + 1) \sqrt{3 m + 1}}$

$\Rightarrow \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} \leq \sqrt{\frac{{(2 m + 1)}^{2}}{4 {(m + 1)}^{2} (3 m + 1)}} \Rightarrow \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} \leq \sqrt{\frac{(4 m^{2} + 4 m + 1) \times (3 m + 4)}{4 (3 m^{3} + 7 m^{2} + 5 m + 1) (3 m + 4)}} \Rightarrow \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} \leq \sqrt{\frac{12 m^{3} + 28 m^{2} + 19 m + 4}{(12 m^{3} + 28 m^{2} + 20 m + 4) (3 m + 4)}} ∵ \frac{12 m^{3} + 28 m^{2} + 19 m + 4}{(12 m^{3} + 28 m^{2} + 20 m + 4)} < 1 ∴ \frac{(2 m + 2)!}{2^{2 m + 2} {((m + 1)!)}^{2}} < \frac{1}{\sqrt{3 m + 4}}$

Thus, P(m + 1) is true.

Hence, by mathematical induction $\frac{(2 n)!}{2^{2 n} (n!)^{2}} \leq \frac{1}{\sqrt{3 n + 1}}$ is true for all n ∈ N

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