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We assume that $\sum |a_n|^{2}$ converges, then I want to conclude that $\prod (1+a_n)$ converges to a non zero element $\iff$ the series $\sum a_n$ converges.

My attempt

If $\prod (1+a_n)$ converges to a non zero element then we can write

$$\prod (1+a_n)= \prod \exp(\ln(1+a_n))= \exp(\sum \ln (1+a_n)) \le \exp(\sum |\ln (1+a_n)|)$$

Then using that $\sum |a_n|^{2}$ converges we can choose $n$ such that $|a_n|^2 < \frac{1}{4}$ and we know that $|ln(1+z)|\le 2 |z|$ if $|z|< \frac{1}{2}$ using the series expansion, we get that

$$\exp(\sum |\ln (1+a_n)|) \le \exp(\sum 2|a_n|)$$

For the converse I want to use the result that says that if $\sum |a_n|$ converges then $\prod (1+a_n)$ converges so since we have that $\sum |a_n|^{2}$ converges then $\sum |a_n|$ converges and $\prod (1+a_n)$ does too.

Questions

But from here I don't know if I am right, how to conclude and solve the converse part to say that we have a non zero limit, and another thing Can someone provide explicit examples of a sequence of complex numbers $\{a_n\}$ such that $\sum a_n$ converges but $\prod (1+a_n)$ diverges and the other way around (This is $\prod (1+a_n)$ converges but $\sum a_n$ diverges )?

Thanks a lot in advance.

RRL
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user162343
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  • The product $\prod (1 + a_n)$ and the series $\sum a_n$ must converge or diverge together when $a_n \in \mathbb{R}$, but only for the separate cases where $a_n \geqslant 0$ and $-1 < a_n < 0$ for all $n$. If $a_n$ changes sign periodically you can find cases where the product and series are not simultaneously convergent (even when the terms are real). For examples see http://math.stackexchange.com/q/1679365/148510 – RRL Apr 13 '16 at 14:18
  • Let me check your reference but could you provide a detailed proof of the other facts ? – user162343 Apr 13 '16 at 14:23
  • Related: http://math.stackexchange.com/questions/349713/suma-n2-infty-and-sum-a-n-infty-implies-sum-log1a-n-converges – Martin Sleziak Apr 13 '16 at 14:28
  • Let me check it just a moment, Can I let you know if there is something missing or questions ? – user162343 Apr 13 '16 at 14:36
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    @RRL How do you cocnlude that you don't have a zero limit in the product? – user162343 Apr 14 '16 at 13:59

2 Answers2

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If $\sum |a_n|^2$ converges, then $|a_n| \to 0$ and there is a positive integer $N$ such that $|a_n| < 1/2$ for all $n > N$.

We then have

$$| \log(1 + a_n) - a_n| = \left|\sum_{k=2}^\infty (-1)^{k+1}\frac{a_n^k}{k} \right| \leqslant \sum_{k=2}^\infty \frac{|a_n|^k}{k} \leqslant \frac1{2}|a_n|^2 \sum_{k=0}^\infty |a_n|^k \\ = \frac{|a_n|^2}{2(1 - |a_n|)} < |a_n|^2 $$

Hence, the series $\sum (\log(1+a_n) - a_n)$ is absolutely convergent. Therefore, $\sum a_n$ and $\sum \log(1+a_n)$ converge or diverge together , which implies that $\sum a_n$ and $\prod (1+a_n)$ converge or diverge together

RRL
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  • Let me check it but this proves both sides right ? – user162343 Apr 13 '16 at 17:23
  • It does prove both sides. For your converse use $\sum \log (1 + a_n) = \sum ( \log( 1+ a_n) - a_n) + \sum a_n$. Then convergence of $\sum a_n$ implies convergence of $\sum \log(1 + a_n)$, since both series on the RHS converge, – RRL Apr 13 '16 at 17:28
  • ... and the other implication follows from $\sum a_n = \sum ( \log( 1+ a_n) - a_n) - \sum \log( 1+ a_n)$ – RRL Apr 13 '16 at 17:57
  • How do you conclude that is a non-zero limit (in the product)? – user162343 Apr 14 '16 at 13:43
  • The product has a zero limit if and only if $\sum \log(1 + a_n)$ diverges to $- \infty$. So if $\sum a_n$ is convergent that will not happen. – RRL Apr 14 '16 at 14:07
  • Also the product could be zero if for one of the terms $a_n = -1$. If this occurs for finitely many $n$ then you have a degenerate case and you consider what happens to the product after the last point at which a term is zero. This is usually excluded in the hypotheses of the theorem. – RRL Apr 14 '16 at 14:10
  • Ok, but I didn't understand your last comment :) – user162343 Apr 14 '16 at 14:12
  • I'm just excluding the trivial case where $\log(1 + a_n)$ is not defined. One hypothesis of this theorem should be $a_n > -1$ for all $n$. – RRL Apr 14 '16 at 14:23
  • aaa ok so you think that the problem is missing that condition? But I can argue as you did right, we know that we have a zero limit if the logarithm series diverges, then by your above proof we have that no and the product is well defined right? – user162343 Apr 14 '16 at 14:26
  • Correct. I'm sure the condition $a_n > -1$ is implicit. That is almost always stated up front but obviously without it, the product is zero or oscillatory. – RRL Apr 14 '16 at 14:52
  • @RRL I struggled with the above question (an excercise in Stein Shakarachi) for sometime and later proved it exactly the same way as you did, the main clever part with the proof is that we must consider $\log(1+a_n)-a_n$ instead of let us say $\log(1+a_n)$ to exploit the absolute convergence of $|a_n|^2$. Beautifully well written proof. – Sundara Narasimhan May 19 '20 at 16:10
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You're on the right track. You've said that for small $x$, $\ln(1+x) < 2x$. That gave you an upper bound.

Can you make a similar claim, such as "for small enough $x$, $\ln(1 + x) > x/2$", for example? If so, you can reverse all your inequalities and do the lower bound proof.

I'm not saying that the bound I've suggested above is correct...merely that it might be, and you should try to prove it or find something similar that works.

John Hughes
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  • Well I have found http://math.stackexchange.com/questions/234043/infinite-product-convergence-problem?rq=1 but I don't know if it is what we want :) – user162343 Apr 13 '16 at 13:25
  • And what the examples and the converse part? – user162343 Apr 13 '16 at 13:26
  • The point of suggesting a lower bound was to help with the converse part. For the examples: to find such a thing, you'd need a sequence where the sum of the squared moduli is NOT bounded. (Because if it IS bounded, then the two notions are equivalent, as you'll have just proved!) That might help you look for examples in the right places. Also: why do you think that there have to be examples for both directions? Perhaps under ALL conditions, we have one implication, but only under the bounded squared moduli sum condition do we have the other. If so, you should expect only one example. – John Hughes Apr 13 '16 at 13:39
  • Let me provide you a further result, just a second. – user162343 Apr 13 '16 at 13:50
  • Ready I have edited my post so I hope that we have two examples. – user162343 Apr 13 '16 at 13:55
  • Remember we are in complex analysis :) . – user162343 Apr 13 '16 at 14:00