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Show that mathematical induction can be used to prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker inequality we originally tried to prove using mathematical induction.

Base Case: P(2) \begin{aligned} \frac{1}{2}\cdot\frac{2(2)-1}{2(2)} &< \frac{1}{\sqrt{3(2) + 1}}\\ \frac{3}{8} &< \frac{1}{\sqrt{7}}\\ \frac{1}{8} &< \frac{1}{3\sqrt{7}}\\ \end{aligned}

This is true as $8 > 3\sqrt{7}$.

Inductive Hypothesis: $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}}$

In the inductive step, we want to show that $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n+4}}$.

Using the inductive hypothesis, we can get to the following:

\begin{aligned} \frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} &< \frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\ &< \frac{1}{\sqrt{3n+1}}\cdot 1\\ \end{aligned}

I am not sure how to get to $< \frac{1}{\sqrt{3n+4}}$ from here because i know that if the denominator would get bigger by adding 3 to it so the inequality wouldn't follow...

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    $\begingroup$ Using the same techniques, a lower bound of $\frac1{2\sqrt{n}}$ can be shown. For comparison, using the Gamma Function, we get the bounds $\frac1{\sqrt{\pi(n+1/2)}}\le\frac12\frac34\cdots\frac{2n-1}{2n}\le\frac1{\sqrt{\pi n}}$. $\endgroup$
    – robjohn
    Oct 22, 2018 at 8:56

3 Answers 3

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Continuing where you left. You need to show $\large{\frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}<\frac{1}{\sqrt{3n+4}}}$, as square roots defined as positive numbers we can multiply both side by $\sqrt{3n+1},$ and this implies $$\frac{2n+1}{2n+2}< \frac{\sqrt{3n+1}}{\sqrt{3n+4}}\implies (2n+1)\sqrt{3n+4}<(2n+2)\sqrt{3n+1}\\ \implies (4n^2+4n+1)(3n+4)<(4n^2+8n+4)(3n+1)\\ \implies(4n^2+4n+1)\big(3n+4-3n-1)<(4n+3)(3n+1)\\ \implies 12n^2+12n+3<12n^2+13n+3 $$ which is true. Hence we are done.

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  • $\begingroup$ Thank you! When you start off proving what you need to show, how do you know that $(2n+1)\sqrt{3n+4} < (2n+2)\sqrt{3n+1}$? In other words, how do you know that the offset of 3 in the sqrt is going to be > the offset of 1 in the multiplication on the other side $\endgroup$
    – banana
    Oct 22, 2018 at 3:41
  • $\begingroup$ @JerseyFonseca It is not known at the begining. Let me edit it. $\endgroup$ Oct 22, 2018 at 3:43
  • $\begingroup$ It was a try. If you ask why I have tried to show it, the answer is in the last line on the solution. $\endgroup$ Oct 22, 2018 at 3:44
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    $\begingroup$ Your implications are backwards, and as such they do not show what you want. Actually, they are biconditionals, and that is what would make it work. $\endgroup$
    – robjohn
    Oct 22, 2018 at 8:50
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Assuming $n\ge0$, we can show that $$ \frac{2n+1}{2n+2}\le\frac{\sqrt{3n+1}}{\sqrt{3n+4}}\tag1 $$ by squaring both sides to get the equivalent $$ \frac{4n^2+4n+1}{4n^2+8n+4}\le\frac{3n+1}{3n+4}\tag2 $$ and cross-multiplying to get the equivalent $$ 12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4\tag3 $$ which is true since $n\ge0$.


Therefore, if $$ \frac12\frac34\cdots\frac{2n-1}{2n}\le\frac1{\sqrt{3n+1}}\tag4 $$ applying $(1)$, we get $$ \begin{align} \frac12\frac34\cdots\frac{2n-1}{2n}\color{#C00}{\frac{2n+1}{2n+2}} &\le\frac1{\sqrt{3n+1}}\color{#C00}{\frac{\sqrt{3n+1}}{\sqrt{3n+4}}}\\ &=\frac1{\sqrt{3n+4}}\tag5 \end{align} $$

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  • $\begingroup$ @JerseyFonseca Apply this to your second last line. $\endgroup$ Oct 22, 2018 at 3:39
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You need to show that $$\frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\ < \frac{1}{\sqrt{3n+4}}\\$$Square both sides and cross multiply to get $$ (2n+1)^2 (3n+4)<(3n+1)(2n+2)^2$$

Multiply and cancel equal terms to get $$ 12n^2+12n+3 <12n^2 +13n+3 $$

Which is true for all positive integers.

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