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- Feb 14, 2012

- 3,802

- Thread starter anemone
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- Feb 14, 2012

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[tex]\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6\cdots (2n)} \leq \frac{1}{\sqrt{3n+1}}[/tex]

for all positive integers n. The result then follows by setting n = 500000 and observing that

[tex]\frac{1}{\sqrt{1500001}} < \frac{1}{1000}[/tex]

The claim is proved by induction on n.

For n = 1, the claim is obvious.

Assume the claim is true for n. We have to show that

[tex]\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)(2n+1)}{2 \cdot 4 \cdot 6\cdots (2n)(2n+2)} \leq \frac{1}{\sqrt{3n+4}}[/tex]

But

[tex]\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)(2n+1)}{2 \cdot 4 \cdot 6\cdots(2n)(2n+2)}=\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6\cdots (2n)} \frac{2n+1}{2n+2}\leq\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}[/tex]

So we have to show that

[tex]\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+4}}[/tex]

This inequality follows by clearing fractions, squaring both sides and simplifying. The result is

[tex]19n \leq 20n[/tex]

which holds for all positive n. This completes the proof of the claim.

- Feb 13, 2012

- 1,704

http://mathhelpboards.com/questions...mber-theory-other-sites-7479-4.html#post40097

... the the explicit expression of the sequence is...

$\displaystyle a_{n} = \prod_{k=1}^{n} (1 - \frac{1}{2\ k})\ (1)$

... and it is the solution of the difference equation...

$\displaystyle a_{n+1} = a_{n}\ (1 - \frac{1}{2\ n}),\ a_{1}=1\ (2)$

The (2) is related to the ODE...

$\displaystyle y^{\ '} = - \frac{y}{2\ x}\ (3)$

... the solution of which is $\displaystyle y = \frac{c}{\sqrt{x}}$, so that we can suppose $\displaystyle a_{n} \sim r_{n}= \frac{c}{\sqrt{n}}$. If we suppose that $r_{500000}= \frac{1}{1000}$ then is $\displaystyle c = \frac{1}{\sqrt{2}}$. In the following table the first values os $a_{n}$ and $r_{n}$ are reported...

$a_{1}= 1,\ r_{1} = .70710678...$

$a_{2}= .5,\ r_{2} = .5$

$a_{3}= .375,\ r_{3} = .40824829...$

$a_{4}= .3125,\ r_{4} = .35355339...$

$a_{5}= .273438...,\ r_{5} = .31627766...$

$a_{6}= .246094...,\ r_{6} = .2886751...$

$a_{7}= .225586...,\ r_{7} = .2672612...$

$a_{8}= .209473...,\ r_{8} = .25$

$a_{9}= .196381...,\ r_{9} = .2357022...$

$a_{10}= .185471...,\ r_{10} = .2236067...$

It is clear from the table that for n 'large enough' the relative increments of the $a_{n}$ and $r_{n}$ are pratically the same and that is verified considering that is...

$\displaystyle \frac{a_{n+1}}{a_{n}} = 1 - \frac{1}{2\ n}$

$\displaystyle \frac{r_{n+1}}{r_{n}} = \sqrt{1 - \frac{1}{n}} = 1 - \frac{1}{2\ n} - \frac{1}{8\ n^{2}} - ...\ (4)$

... so that we can conclude that is $\displaystyle a_{500000} < r_{500000} = \frac{1}{1000}$...

Kind regards

$\chi$ $\sigma$

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- Feb 14, 2012

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Thanks for participating to both of you, **Petek** and **chisigma**! Your induction method looks nice and great, **Petek**!

@**chisigma**, your solution post reminds me of this thread(http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/find-a_{100000}-8448.html)! Bravo, **chisigma**!

Solution provided by other:

Let $x=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6} \cdots\dfrac{999999}{1000000}$.

Thus, what we need to show is that $x<\dfrac{1}{1000}$.

Now, note that

$x^2=\dfrac{1^2}{2^2}\cdot\dfrac{3^2}{4^2}\cdot \dfrac{5^2}{6^2} \cdots\dfrac{999999^2}{1000000^2}$

Since decreasing the denominator of a fraction makes it bigger, we have that

$\dfrac{1^2}{2^2}\le \dfrac{1^2}{2^2-1}= \dfrac{1^2}{(2-1)(2+1)}=\dfrac{1^2}{1\cdot3}$

$\dfrac{3^2}{4^2}\le \dfrac{3^2}{4^2-1}= \dfrac{3^2}{(4-1)(4+1)}=\dfrac{3^2}{3\cdot5}$

$\dfrac{5^2}{6^2}\le \dfrac{5^2}{6^2-1}= \dfrac{5^2}{(6-1)(6+1)}=\dfrac{5^2}{5\cdot7}$

$\vdots\;\;\;\;\;\;\;\;\;\;\;\vdots$

$\dfrac{999999^2}{1000000^2}\le \dfrac{999999^2}{1000000^2-1}= \dfrac{999999^2}{(1000000-1)(1000000+1)}=\dfrac{999999^2}{999999\cdot1000001}$

Multiplying all these together we get

$x^2<\dfrac{1}{1000001}<\dfrac{1}{1000000}$

Now, taking square roof of both sides we obtain

$x<\dfrac{1}{1000}$ or

$\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6} \cdots\dfrac{999999}{1000000}<\dfrac{1}{1000}$ (Q.E.D.)

@

Solution provided by other:

Thus, what we need to show is that $x<\dfrac{1}{1000}$.

Now, note that

$x^2=\dfrac{1^2}{2^2}\cdot\dfrac{3^2}{4^2}\cdot \dfrac{5^2}{6^2} \cdots\dfrac{999999^2}{1000000^2}$

Since decreasing the denominator of a fraction makes it bigger, we have that

$\dfrac{1^2}{2^2}\le \dfrac{1^2}{2^2-1}= \dfrac{1^2}{(2-1)(2+1)}=\dfrac{1^2}{1\cdot3}$

$\dfrac{3^2}{4^2}\le \dfrac{3^2}{4^2-1}= \dfrac{3^2}{(4-1)(4+1)}=\dfrac{3^2}{3\cdot5}$

$\dfrac{5^2}{6^2}\le \dfrac{5^2}{6^2-1}= \dfrac{5^2}{(6-1)(6+1)}=\dfrac{5^2}{5\cdot7}$

$\vdots\;\;\;\;\;\;\;\;\;\;\;\vdots$

$\dfrac{999999^2}{1000000^2}\le \dfrac{999999^2}{1000000^2-1}= \dfrac{999999^2}{(1000000-1)(1000000+1)}=\dfrac{999999^2}{999999\cdot1000001}$

Multiplying all these together we get

$x^2<\dfrac{1}{1000001}<\dfrac{1}{1000000}$

Now, taking square roof of both sides we obtain

$x<\dfrac{1}{1000}$ or

$\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6} \cdots\dfrac{999999}{1000000}<\dfrac{1}{1000}$ (Q.E.D.)

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