Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me prove by induction that $$ n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1) $$

share|improve this question
1  
This sounds like homework, how far did you get? –  Listing Mar 15 '12 at 10:24
    
When $n=1$, LHS$=1^1=1$ and RHS$=1\cdots(2\cdot1-1)=1$=LHS. So I think you want to prove the inequality when $n\geq 2$. –  Paul Mar 15 '12 at 10:32

7 Answers 7

up vote 10 down vote accepted

Hint:

$$\rm (n+1)^{n+1}>(2n+1)n^n$$

$$\large\Updownarrow $$

$$\rm \left(1+\frac{1}{n}\right)^n>\binom{n}{0}+\binom{n}{1}\frac{1}{n}=2>\frac{2n+1}{n+1}$$

share|improve this answer

This is proof without induction. $$ 1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)= \frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}= \frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot\ldots\cdot (2\cdot n)}= $$ $$ \frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{2^n\cdot 1\cdot 2\cdot 3\cdot\ldots\cdot n}= \frac{(n+1)\cdot (n+2)\cdot\ldots\cdot 2n}{2^n} $$ So we had to prove that $$ n^n>\frac{(n+1)\cdot (n+2)\cdot\ldots\cdot 2n}{2^n} $$ But the last inequality is obvious, since $$ \frac{(n+1)\cdot (n+2)\cdot\ldots\cdot 2n}{2^n}= \frac{n+1}{2}\frac{n+2}{2}\cdots\frac{n+n}{2}<n\cdot n\cdot\ldots \cdot n=n^n $$

share|improve this answer

This could be an interesting method. For k>1 and even n>0, show that

$n^{2k}>[n-(2k-1)]\times[n-(2k-3)]\times...(n-1)\times(n+1)\times...\times[n+(2k-3)]\times[n+(2k-1)]$

Our base case: $n^2>n^2-1=(n-1)\times(n+1)$

Now the inductive step: Assume true for $k=m$, prove true for $k=m+1$

$n^{2m}>[n-(2m-1)]\times[n-(2m-3)]\times...\times[n+(2m-3)]\times[n+(2m-1)]$

$n^2>n^2-(2m+1)^2=[n-(2m+1)]\times[n+(2m+1)]$

Multiply the top inequality by $[n-(2m+1)]\times[n+(2m+1)]$ and the bottom by $n^{2m}$ to get

$n^{2(m+1)}>[n^2-(2m+1)^2]\times n^{2m}>[n-(2m+1)]\times[n-(2m-1)]\times...\times[n+(2m-1)]\times[n+(2m+1)]$.

This completes the inductive step and we have proven our case for k>1. This includes the case $k=\frac n2$.

The odd case should be similar, except you will likely need to actually use that $n>0$ as you will likely need to multiply or divide by odd powers of n which would reverse the inequalities if $n<0$.

share|improve this answer

We need to prove $1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)< n^n$

For $n=1$ we have $1 < 1$, what is wrong.

For $n=2$ we have $3 < 4$ what is true.

Note that $1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1) (2\cdot (n+1)-1)=n^n(2(n+1)-1)$

then prove $(n+1)^{n+1} - n^n\cdot (2\cdot (n+1)-1) > 0$

share|improve this answer

Hint $\ $ Use telescopy: a trivial induction shows a product of terms $> 1$ is $> 1$. Applying this to

$$\rm\ f(n)\ =\ \frac{f(n)}{\color{red}{f(n-1)}}\ \frac{\color{red}{f(n-1)}}{\color{green}{f(n-2)}}\ \frac{\color{green}{f(n-2)}}{\cdots }\ \cdots\ \frac{\cdots}{\color{brown}{f(3)}}\ \frac{\color{brown}{f(3)}}{\color{blue}{f(2)}}\ \frac{\color{blue}{f(2)}}{1}$$

the product $\rm\:f(n) > 1\:$ if each factor is, i.e. if $\rm\:f(2) > 1\:$ and $\rm\:f(k+1)/f(k) > 1\:$ for $\rm\:k\ge 2.\:$ But your $\rm\:f(n) = n^n/(1\cdot 3\cdot 5\cdots 2n\!-\!1)\:$ satisfies such by an easy calculation (see anon's answer).

share|improve this answer

$$ \frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac{1}{n}\right)^n\ge(n+1)\left(1+n\cdot\frac{1}{n}\right)=2n+2>2n+1 $$

share|improve this answer

Alternative solution:(Without induction)

Using $AM \ge GM$ inequality we get $$\dfrac{1+3+5+\dots+(2n-1)}{n}> [1\cdot3 \cdot 5\dots (2n-1)]^{1/n}$$ Note that $$1+3+5+\dots+(2n-1) = \sum_{i=1}^n(2i-1)\\=2 \sum_{i=1}^n i-\sum_{i=1}^n1 \\=2\cdot\dfrac{n(n+1)}{2}-n =n^2$$ So our inequality becomes $$\dfrac{n^2}{n}> [1\cdot3 \cdot 5\dots (2n-1)]^{1/n} \\ \implies n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.