# Infinite product

#### Dinkydoe

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.

$$\displaystyle \Rightarrow$$

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?

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#### chisigma

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

Kind regards

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

#### Dinkydoe

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
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.

$$\displaystyle \Rightarrow$$

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 $$\displaystyle N\to\infty$$.

• Dinkydoe