# Help with this proof by induction with inequalities.

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...

• 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 $\frac1{\sqrt{\pi(n+1/2)}}\le\frac12\frac34\cdots\frac{2n-1}{2n}\le\frac1{\sqrt{\pi n}}$.
– robjohn
Oct 22, 2018 at 8:56

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.

• 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 Oct 22, 2018 at 3:41
• @JerseyFonseca It is not known at the begining. Let me edit it. Oct 22, 2018 at 3:43
• It was a try. If you ask why I have tried to show it, the answer is in the last line on the solution. Oct 22, 2018 at 3:44
• 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.
– robjohn
Oct 22, 2018 at 8:50

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}

• @JerseyFonseca Apply this to your second last line. Oct 22, 2018 at 3:39

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.