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Prove by induction for

$$ \frac{\left(1\right)\left(3\right)\left(5\right)...\left(2n-1\right)}{\left(2\right)\left(4\right)\left(6\right)...\left(2n\right)}\le\frac{1}{\sqrt{n+1}}\quad \text{for all integers}\quad n \ge 1$$

My current inductive step: $$ \frac{(2(k+1))!}{2^{2\left(k+1\right)}((k+1)!)^2} = \frac{2k+1}{2(k+1)} \cdot\frac{(2k)!}{2^{2k}\left(k!\right)^2}$$

I'm not sure how to proceed to show $$ \frac{2k+1}{2(k+1)} \cdot \frac{(2k)!}{2^{2k}\left(k!\right)^2} \le \frac{1}{\sqrt{k+2}}$$

Martin Sleziak
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j_remeh
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1 Answers1

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There's no need to use factorials here, you can do it with basic arithmetic. The base case is

$$\frac{1}{2} \leq \frac{1}{\sqrt{2}},$$

which you can square both sides to find that it's true.

Then we note that if $f(n)$ is the fraction on the left of the inequality, then $f(n+1) = f(n)\frac{2n+1}{2n+2}$.

So we assume for induction that $f(n) \leq \frac{1}{\sqrt{n+1}}$, then:

$$f(n) \leq \frac{1}{\sqrt{n+1}}$$ $$f(n)\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{n+1}}\frac{2n+1}{2n+2}$$ $$f(n+1) \leq \frac{2n+1}{\sqrt{n+1}(2n+2)}$$

Now we want to know if

$$\frac{2n+1}{\sqrt{n+1}(2n+2)} \leq \frac{1}{\sqrt{n+2}}$$ $$(2n+1)\sqrt{n+2} \leq \sqrt{n+1}(2n+2)$$ $$(2n+1)^2(n+2) \leq (n+1)(2n+2)^2$$ $$4 n^3 + 12 n^2 + 9 n + 2 \leq 4 n^3 + 12 n^2 + 12 n + 4$$ $$9 n \leq 12 n + 2$$

Which is easy to see to be true for any positive $n$.

orlp
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