Prove that (2n)! 2^ 2n (n!)^2 1 3n+1 for all n .

Question

Prove that $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$ for all $\text{n}\in\text{N}.$

✓

Answer

P(n): $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$
For n = 1
$\frac{2!}{2^2.1}\leq\frac{1}{\sqrt{4}}$
$=\frac{1}{2}\leq\frac{1}{2}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\frac{(2\text{k})!}{2^{2\text{k}}(\text{k}!)^2}\leq\frac{1}{\sqrt{3\text{k}+1}} \ ...(1)$
We have to show that,
$\frac{2(\text{k+1})!}{2^{2(\text{k+1})}\big[(\text{k+1})!\big]^2}\leq\frac{1}{\sqrt{3\text{k}+4}}$
Now,
$\frac{2(\text{k+1})!}{2^{2(\text{k+1})}\big[(\text{k+1})!\big]^2}$
$=\frac{(2\text{k+2})!}{2^{2\text{k}}.2^{2}(\text{k+1})!{(\text{k+1})!}}$
$=\frac{(2\text{k+2})(2\text{k+1})(2\text{k})!}{4.2^{2}(\text{k+1})(\text{k}!){(\text{k+1})(\text{k}!)}}$
$=\frac{2(\text{k+2})(2\text{k+1})(2\text{k})!}{4.(\text{k+1})^2.2^{2\text{k}}.(\text{k}!)}^2$
$\leq\frac{2(2\text{k+1})}{4(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+1}}$
$\leq\frac{(2\text{k+1})}{2(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+1}}$
$\leq\frac{(2\text{k+1})}{2(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+3+1}}$
$\leq\frac{1}{\sqrt{3\text{k}+4}} \ \begin{bmatrix}\text{Since,} \ 2 \text{k}+2<2\text{k}+2\\\ 3\text{k}+1\leq 3\text{k}+4\end{bmatrix}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI

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