Infinite product

**Dinkydoe** · May 8th 2010, 09:27 AM

A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.

If $b_n>1$ for all $n$ , then prove that $\prod_n b_n$ converges iff $\sum_{n}\log(b_n) < \infty$
I'm used to not so easy homework-assignments, so that's why I feel I'm missing something here.

My answer would be:

$\Rightarrow$

if $\prod_n b_n=M$ we have $M\geq 1$ since $b_n>1$ for all $n$ . Hence $\log(M)<\infty$ and thus we obtain $\log(M)=\log(\prod_n b_n)= \sum_n \log(b_n) <\infty$

$\Leftarrow$
If $\sum_n \log(b_n) = \log(\prod_n b_n) = M$ we have $e^M= \prod_n b_n<\infty$

So am I missing something?

**chisigma** · May 8th 2010, 09:34 AM

A more general criterion of convergence is that $b_{n} >0 , \forall n$ and $\sum_{n} \ln (b_{n})$ converges...

Kind regards

$\chi$ $\sigma$

**Dinkydoe** · May 8th 2010, 09:47 AM

A more general criterion of convergence is that and converges...

I made 2 typo's i saw...i wrote n instead of $b_n$

However, what exactly am I supposed to show here? And what did I miss?
I feel my argument is incomplete, so what is left to show?

**Opalg** · May 8th 2010, 11:14 AM

Originally Posted by Dinkydoe

A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.

If $b_n>1$ for all $n$ , then prove that $\prod_n b_n$ converges iff $\sum_{n}\log(b_n) < \infty$
I'm used to not so easy homework-assignments, so that's why I feel I'm missing something here.

My answer would be:

$\Rightarrow$

if $\prod_n b_n=M$ we have $M\geq 1$ since $b_n>1$ for all $n$ . Hence $\log(M)<\infty$ and thus we obtain $\log(M)=\log(\prod_n b_n)= \sum_n \log(b_n) <\infty$

$\Leftarrow$
If $\sum_n \log(b_n) = \log(\prod_n b_n) = M$ we have $e^M= \prod_n b_n<\infty$

So am I missing something?

Go back to the definitions. To say that $\prod_n b_n$ converges means that $\lim_{N\to\infty}\prod_{n=1}^N b_n$ exists (and is nonzero). To say that $\sum_{n}\log(b_n)$ converges means that $\lim_{N\to\infty}\sum_{n=1}^N \log(b_n)$ exists. Also, if $b_n>1$ for all n then $\sum_n\log(b_n)$ is a series of positive terms, so it converges iff its sum is finite.

Now use the fact that the logarithm function and its inverse are continuous to argue that $\sum_{n=1}^N \log(b_n) = \log\Bigl(\prod_{n=1}^N b_n\Bigr)$ converges iff $\prod_{n=1}^N b_n$ converges, as $N\to\infty$ .

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