# Infinite product converges if and only if sum converges

1. Dec 2, 2013

### ToNoAvail27

1. The problem statement, all variables and given/known data
$a_n$ is a sequence of positive numbers. Prove that $\prod_{n=1}^{\infty} (1+a_n)$ converges if and only if $\sum_{n=1}^{\infty} a_n$ converges.

2. Relevant equations

3. The attempt at a solution
I first tried writing out a partial product: $\prod_{n=1}^{N} (1+a_n) = (1+a_1)(1+a_2)\dots(1+a_N) = 1 + \prod_{n=1}^{N} a_n + \sum_{n=1}^{N} a_n + C$, where $C$ is the sum of all the combinations of the $a_i$, such as $a_1a_2a_N$ etc, but that did not really lead me to much. I was given the hint of using the logarithm, but I am not really sure when I would use that. Perhaps when the sum converges, so does $\log(1+a_n)$, though I'm not sure how that relates.

2. Dec 2, 2013

### jbunniii

How does $\log(1+a_n)$ compare with $a_n$?

3. Dec 2, 2013

### ToNoAvail27

Well, $\log(1+a_n) < a_n$ so it will converge when $a_n$ does?
Also, if it does converge then wouldn't
$\sum_{n=1}^{N} \log(1+a_n) = \log(1+a_1)+\log(1+a_2)+\ldots+\log(1+a_n) = \log(\prod_{n=1}^{N}(1+a_n))$ imply that the product would converge?

4. Dec 2, 2013

### jbunniii

Yes, that's right. The function $\log$ is continuous, which justifies this step:
$$\log \lim_{N \rightarrow \infty} \prod_{n=1}^{N} (1+a_n) = \lim_{N \rightarrow \infty} \log \prod_{n=1}^{N} (1+a_n)$$
Then apply the product-to-sum property of the log to get
$$\log \prod_{n=1}^{N} (1+a_n) = \sum_{n=1}^{N} \log (1+a_n)$$
The right hand side is smaller than $\sum_{n=1}^{N} a_n$. Putting it all together and taking limits gives you a proof that $\log \prod_{n=1}^{\infty} (1+a_n) \leq \sum_{n=1}^{\infty}a_n$. What can you conclude?

Note that the problem statement says "if and only if", so you still need to prove the implication in the other direction.

5. Dec 2, 2013

### Ray Vickson

For 0 < x < 1 we have
$$x -\frac{1}{2} x^2 < \ln(1+x) < x$$
This gives you valuable information when N is so large that $a_n < 1 \: \forall n \geq N$.

6. Dec 2, 2013

### Ray Vickson

For 0 < x < 1 we have
$$x -\frac{1}{2} x^2 < \ln(1+x) < x$$
This gives you valuable information when n is so large that $a_n < 1$.