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Here is my problem:

Use induction to show that for $n\ge1$, $$\frac{4^n}{\sqrt{4n}} \le \frac{(2n)!}{(n!)^2}\le \frac{4^n}{\sqrt{3n+1}}.$$

What I have so far:

For $n=1$ it is $2≤2≤2$, and I tried to originally do this problem by assuming that the case holds for $n\ge 1$, and if it hold for n+1, than it should hold for all $n\ge 1$.

I started with $n+1$ case: $$\frac{4^{n+1}}{\sqrt{4(n+1)}}\le\frac{(2(n+1))!}{((n+1)!)^2}\le\frac{4^{n+1}}{\sqrt{3(n+1)+1}},$$ but this did not really lead me anywhere.

Ѕᴀᴀᴅ
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SBob
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1 Answers1

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Note that$$\frac{\frac{\bigl(2(n+1)\bigr)!}{(n+1)!^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)(2n+1)}{(n+1)^2}=2\frac{2n+1}{n+1}.$$On the other hand$$\frac{\frac{4^{n+1}}{\sqrt{4(n+1)}}}{\frac{4^n}{\sqrt{4n}}}=2\sqrt{\frac n{n+1}}$$and$$\frac{\frac{4^{n+1}}{\sqrt{3n+4}}}{\frac{4^n}{\sqrt{3n+1}}}=4\sqrt{\frac{3n+1}{3n+4}}.$$Can you take it from here?

José Carlos Santos
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