Infinite product

Dec 2009
411
131
A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.


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

My answer would be:

\(\displaystyle \Rightarrow\)

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

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

So am I missing something?
 
Last edited:

chisigma

MHF Hall of Honor
Mar 2009
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994
near Piacenza (Italy)
A more general criterion of convergence is that \(\displaystyle b_{n} >0 , \forall n\) and \(\displaystyle \sum_{n} \ln (b_{n})\) converges...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Dec 2009
411
131
A more general criterion of convergence is that and converges...
I made 2 typo's i saw...i wrote n instead of \(\displaystyle b_n\)

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

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.


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

My answer would be:

\(\displaystyle \Rightarrow\)

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

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

So am I missing something?
Go back to the definitions. To say that \(\displaystyle \prod_n b_n\) converges means that \(\displaystyle \lim_{N\to\infty}\prod_{n=1}^N b_n\) exists (and is nonzero). To say that \(\displaystyle \sum_{n}\log(b_n)\) converges means that \(\displaystyle \lim_{N\to\infty}\sum_{n=1}^N \log(b_n)\) exists. Also, if \(\displaystyle b_n>1\) for all n then \(\displaystyle \sum_n\log(b_n)\) 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 \(\displaystyle \sum_{n=1}^N \log(b_n) = \log\Bigl(\prod_{n=1}^N b_n\Bigr)\) converges iff \(\displaystyle \prod_{n=1}^N b_n\) converges, as \(\displaystyle N\to\infty\).
 
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