Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ .

Question

Prove $$ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $$

Can this be done by induction using the pi function. If no, why not.

Almost a duplicate: http://math.stackexchange.com/questions/1899857/mathematical-problem-induction/ — StubbornAtom, Sep 25 '16 at 13:16
@user371530: what the prime counting function has to do with the given inequality? — Jack D'Aurizio, Sep 25 '16 at 17:34
$$\prod_{i=1}^n ( {2n-1/2n} ) < ( {1/{2n-1}^{0.5}})$$
Using this, Can it be proved by induction, 1: Put ${n=1}$ . Show that it holds true. 2. Show that it holds true for ${n=k}$ . 3. Prove its true for ${n= k+1}$.

Can we use that algorithm here (using the above function) ? — user371530, Sep 27 '16 at 03:04

score 7 · Answer 1 · answered Sep 25 '16 at 05:57

7

Let $$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}$$ $$B=\frac23\cdot\frac45\cdot\frac67\cdot...\frac{2n}{2n+1}$$ Then $$A<B$$ Then $$A^2<AB=\frac1{2n+1}$$ $$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$$

answered Sep 25 '16 at 05:57

Roman83

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My idea was this: $$\prod_{i=1}^n {(2n-1)/(2n)} < (1/(2n-1)^{0.5}) $$
Using this, 1: Show $n=1$ holds true (substitution). 2: Show $n=k$holds true (substitution). 3. Prove $n=(k+1)$ is true (Proof). This is the algorithm for induction, Can this be used?
– user371530 Sep 30 '16 at 16:35

score 4 · Answer 2 · answered Sep 25 '16 at 06:00

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Hint. If you want to prove it by induction, you may observe that, for $n\ge1$, $$ (2n+2)^2=4n^2+8n+4>4n^2+5n+3=(2n+1)(2n+3) $$ giving $$ \frac1{2n+2}<\frac1{\sqrt{2n+1}\cdot \sqrt{2n+3}}, \qquad n\ge1, $$ which is equivalent to the inductive step: $$ \frac1{\sqrt{2n+1}}\cdot \frac{2n+1}{2n+2}<\frac1{\sqrt{2n+3}}. $$

answered Sep 25 '16 at 06:00

Olivier Oloa

120,989

Would using Π ( x ) in LHS for induction be an incorrect approach? – user371530 Sep 25 '16 at 08:19
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@user371530 Please I don't know what $\Pi (x)$ means. If you have a link to the definition of $\Pi (\cdot)$ maybe I it will be clearer for me. Thanks. – Olivier Oloa Sep 25 '16 at 08:38

score 3 · Accepted Answer · answered Sep 25 '16 at 18:01

An interesting way may be the following: if we define $$ a_n = \int_{0}^{\pi/2}\sin^n(x)\,dx \tag{1}$$ we clearly have that $\{a_n\}_{n\geq 1}$ is a decreasing sequence.
On the other hand, integration by parts gives: $$ a_{2n} = \frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n},\qquad a_{2n+1}=\frac{1}{2n+1}\cdot\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}\tag{2} $$ and we want to give an upper bound to $$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \binom{2n}{n}\frac{1}{4^n}\tag{3} $$ so we may just exploit $a_{2n+1}>a_{2n+2}$, leading to: $$\frac{1}{2n+1}\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}>\frac{\pi}{2}\binom{2n+2}{n+1}\frac{1}{4^{n+1}}=\frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n}\cdot\frac{2n+1}{2n+2} \tag{4}$$ or to: $$\left(\binom{2n}{n}\frac{1}{4^n}\right)^2< \color{red}{\frac{2}{\pi}\cdot\frac{2n+2}{(2n+1)^2}}\tag{5}$$ that is much stronger than the wanted inequality.

Thats a great solution. My idea was this: $$\prod_{i=1}^n ( {(2n-1)/2n} ) < ( {1/({2n-1})^{0.5}})$$ Using this, 1: Show $n=1$ holds true (substitution). 2: Show $n=k$ holds true (substitution). 3. Prove $n=(k+1)$ is true (Proof). This is the algorithm for induction, Can this be used? Thank you for the answer BTW. — user371530, Sep 27 '16 at 04:10

Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ .

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