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I'm reading Analytic Inequalities by Nicholas D. Kazarinoff. On page 5, we are trying to use induction to prove the inequality $$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} < \frac{1}{\sqrt{3n + 1}} . $$ For the inductive step, we want to show that it holds for $n + 1$, i.e. $$ \frac{1}{\sqrt{4n + 5}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 1}{2n} \cdot \frac{2n + 1}{2n + 2} < \frac{1}{\sqrt{3n + 4}} . $$ Kazarinoff says that this is true if $$ \frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}} $$ is also true. I'm trying to figure out why this is the case, because it's not obvious to me.

Edit: this is completely nonsensical because it's circular.

What I've tried:

\begin{align} \frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} &> \frac{1}{\sqrt{4n + 5}} \\ \implies \frac{1}{\sqrt{4n + 1}} &> \frac{1}{\sqrt{4n + 5}} \cdot \frac{2n + 2}{2n + 1} \\ &< \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} \\ &> \frac{1}{\sqrt{4n + 1}} \end{align} That's a pretty useless result that hasn't gotten me anywhere. Any ideas?

Jeremy Lindsay
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2 Answers2

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although an elementary solution is aesthetically preferable (and a good algebraic solution was posted and then for some reason deleted by Soumyadwip Chanda), it may be of interest to see the number-theoretic result posted by OP as a corollary of a more general theorem of real analysis:

suppose that for $a, b \in (0,1)$ we had:

$$ \bigg(\frac{1-a}{1+a}\bigg)^b \lt \frac{1-ba}{1+ba} \tag{1} $$

then, as a special case with $b=\frac12$ and $a = \frac2{4n+3}$, we would have:

$$ \bigg(\frac{1-\frac2{4n+3}}{1+\frac2{4n+3}}\bigg)^{\frac12} \lt \frac{1-\frac1{4n+3}}{1+\frac1{4n+3}} $$

which simplifies to:

$$ \sqrt{\frac{4n+1}{4n+5}} < \frac{2n+1}{2n+2} $$

the formula (1) can be verified by taking the Taylor series for the logarithms of the two sides, which gives:

$$ -2\bigg(ab + \frac{a^3b}3 + \frac{a^5b}5+...\bigg) \lt -2\bigg(ab + \frac{a^3b^3}3 + \frac{a^5b^5}5+...\bigg) $$

David Holden
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By induction assumption, you have $$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n}.$$ Suppose you have $$\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}.$$ Then $$\frac{1}{\sqrt{4n + 5}}<\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2}<\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n}\cdot \frac{2n + 1}{2n + 2}.$$

cqfd
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