where do you suggest i start with this one (series limit)

$\displaystyle \lim_{n\to\infty}\frac{1\cdot 3\cdot 5 \cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \cdots \cdot (2n)}$

i know the limit should be 0 because the the series in the numerator is larger the the one in the denominator for every n

but i don't know how to start on proving it

help?

Re: where do you suggest i start with this one (series limit)

I believe you mean the denominator is larger than the numerator; therefore, the limit is 0. Wouldn't proving $\displaystyle 2n > 2n-1$ (and then pointing out that the product of two larger numbers is larger than the product of two smaller numbers) suffice?

Re: where do you suggest i start with this one (series limit)

Quote:

Originally Posted by

**idom87** $\displaystyle \lim_{n\to\infty}\frac{1\cdot 3\cdot 5 \cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \cdots \cdot (2n)}$

i know the limit should be 0 because the the series in the numerator is larger the the one in the denominator for every n

but i don't know how to start on proving it

help?

You can start writing the limit as...

$\displaystyle \lim_{n \rightarrow \infty} \frac{1\ 3\ 5\ ...\ (2n-1)}{2\ 4\ 6\ ...\ (2n)}= \lim_{n \rightarrow \infty} \prod_{k=1}^{n} (1-\frac{1}{2k})$ (1)

Now the limit (1) is called *infinite product* and a basic criterion exstablishes that if the infinite product is written in the form...

$\displaystyle \prod_{k=1}^{\infty} (1 \pm a_{k})$ (2)

... where all the $\displaystyle a_{k}\ge 0$, then if the series...

$\displaystyle \sum_{k=1}^{\infty} a_{k}$ (3)

... converges, then the infinite product (2) converges and if the (3) diverges, then the infinite product (2) tends to infinity if in (2) the sign is '+' and the infinite product tends to 0 if the sign in (2) is '-'. In Your case the series...

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{2k}$ (4)

... diverges and the sign is '-', so that...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: where do you suggest i start with this one (series limit)

can you do it without using products? i haven't gotten there yet

Re: where do you suggest i start with this one (series limit)

your'e right i mixed that up that's what i meant

Re: where do you suggest i start with this one (series limit)

Quote:

Originally Posted by

**idom87** can you do it without using products? i haven't gotten there yet

All right!... setting...

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

... is...

$\displaystyle \ln p_{n}= \sum_{k=1}^{\infty} \ln (1-\frac{1}{2k})$ (2)

Now if we remember that...

$\displaystyle \ln (1-x)= -x -\frac{x^{2}}{2}-\frac{x^{3}}{3}-...$ (3)

... is also $\displaystyle |\ln(1-x)|>|x|$ so that...

$\displaystyle |\sum_{k=1}^{n} \ln (1-\frac{1}{2k})| > |\sum_{k=1}^{n} \frac{1}{2k}| $(4)

But the series in the right term of (4) diverges and, because for all n is $\displaystyle \ln p_{n}<0$ is...

$\displaystyle \lim_{n \rightarrow \infty} \ln p_{n}= - \infty \implies \lim_{n \rightarrow \infty}p_{n}=0$ (5)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$