Prove that $\frac{1}{1999} < \prod_{i=1}^{999}{\frac{2i−1}{2i}} < \frac{1}{44}$

Question

Prove that $$\dfrac{1}{1999} < \prod_{i=1}^{999}{\dfrac{2i−1}{2i}} < \dfrac{1}{44}$$

from the 1997 Canada National Olympiad.

I have been able to prove the left half of the inequality using induction. Need help with the second part.

---

Prove $\prod_{i=1}^{n}{(1-\frac{1}{2i})} \ge \frac{1}{2n}$

Define $$p(n)=\prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \tag{1}$$

We wish to show $$p(n)\ge\frac{1}{2n},\quad\forall{n}\in\mathbb{Z^+} \tag{2}$$

For the base case, $n=1$, we have $p(1)=1-\frac{1}{2}=\frac{1}{2}\ge\frac{1}{2}$ which holds.

Assume the induction hypothesis (2) for $n$. Then for $n+1$, we can write:

$$\begin{align} p(n+1) = \prod_{i=1}^{n+1}{\left(1-\frac{1}{2i}\right)} &= \left(1-\frac{1}{2(n+1)}\right) \cdot \prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \\ &\ge \left(1-\frac{1}{2(n+1)}\right) \cdot \frac{1}{2n} &(\text{by IH}) \\ &\ge \left(\frac{2n+1}{2(n+1)}\right) \cdot \frac{1}{2n} \\ &\ge \left(\frac{1}{2(n+1)}\right) \cdot \frac{2n+1}{2n} \\ &> \left(\frac{1}{2(n+1)}\right) &(\text{for }n\ge1) \end{align}$$

This proves the induction step, establishing (2) for all positive $n$. Hence, we conclude

$$\boxed{\prod_{i=1}^{999}{\dfrac{2i−1}{2i}} = p(999) \ge \frac{1}{2\times999} = \frac{1}{1998} > \frac{1}{1999}}$$

Prove $\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} < \frac{1}{44}$

For this part, I have applied AM-GM to get an upper bound with a harmonic sum.

$$\left(\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)}\right)^{1/999} \le \frac{1}{999} \cdot \sum_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} \tag{3}$$

I can't see a way of evaluating the summation on the RHS.

Answer 1

You have made the problem much hard than it is suppose to be.

For the left hand side, simply observe that $\frac{1}{2}>\frac{1}{3}$ and $\frac{3}{4}>\frac{3}{5}$ and so on... to $\frac{1997}{1998}>\frac{1997}{1999}$. Now, take the product of these inequalities to get the desired result.

For the other inequality, it is a similar trick except you need to observe that $P^2<\frac{1}{1999}<\frac{1}{1936}=\frac{1}{44^2}$.

Answer 2

$${ P }_{ n }^{ 2 }\left( 999 \right) =\frac { { 1 }^{ 2 }\cdot { 3 }^{ 2 }...\cdot { \left( 2\cdot 999-1 \right) }^{ 2 } }{ { 2 }^{ 2 }\cdot { 4 }^{ 2 }...\cdot { \left( 2\cdot 999 \right) }^{ 2 } } =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\cdot 5 }{ { 4 }^{ 2 } } \cdot ...\cdot \frac { \left( 2\cdot 999-1 \right) \left( 2\cdot 999+1 \right) }{ { \left( 2\cdot 999 \right) }^{ 2 } } \cdot \frac { 1 }{ 2\cdot 999+1 } <\frac { 1 }{ 2\cdot 999+1 } =\frac { 1 }{ 1999 } <\frac { 1 }{ 1936 } =\frac { 1 }{ { 44 }^{ 2 } } $$

Answer 3

Since $\frac{2i-1}{2i}\lt\frac{2i}{2i+1}$, we have $$ \left(\prod_{i=1}^{999}{\frac{2i−1}{2i}}\right)^2\lt\frac14\prod_{i=2}^{999}{\frac{2i−1}{2i}}\prod_{i=2}^{999}{\frac{2i}{2i+1}}=\frac14\frac3{1999}\lt\frac14\frac1{666}\tag1 $$ Since $\frac{2i-1}{2i}\gt\frac{2i-2}{2i-1}$, we have $$ \left(\prod_{i=1}^{999}{\frac{2i−1}{2i}}\right)^2\gt\frac14\prod_{i=2}^{999}{\frac{2i−1}{2i}}\prod_{i=2}^{999}{\frac{2i-2}{2i-1}}=\frac14\frac2{1998}=\frac14\frac1{999}\tag2 $$ Therefore, since $32^2=1024\gt999$ and $25^2=625\lt666$, we have $$ \frac1{64}\lt\frac12\frac1{\sqrt{999}}\lt\prod_{i=1}^{999}{\frac{2i−1}{2i}}\lt\frac12\frac1{\sqrt{666}}\lt\frac1{50}\tag3 $$