Show that if $\sum_{n=1}^{\infty} a_n$ converges conditionally, then $\prod_{n=1}^{\infty} (1+a_n)$ converges conditionally or diverges to 0.

Question

Show that if $\sum_{n=1}^{\infty} a_n$ converges conditionally, then depending on $\sum_{n=1}^{\infty} {a_n}^2$ converges or not, $\prod_{n=1}^{\infty} (1+a_n)$ converges conditionally or diverges to 0.

My thought is that if $\sum_{n=1}^{\infty} {a_n}^2$ converges, then it converges absolutely, and so does $\prod_{n=1}^{\infty} (1-{a_n}^2)=\prod_{n=1}^{\infty} [(1+a_n)(1-a_n)]$. ([ ] here is to emphasize that we treat what's inside as one, not more, items.)

So probably, we have $(1+a_1)(1-a_1)(1+a_2)(1-a_2)\dots$ converges (though not necesarily abosolutely.) Can we proceed from this to $\prod_{n=1}^{\infty} (1+a_n)$ converges? (I guess taking log could possibly work. Edit: I guess log is useful in analysis because it provides a way to make 'parallel' relation between log (or log plus a few powers) and a power, as illustrated in the answer.)

A note: I'm still working on it. This also serves as a record of my progress.

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From the example that $\prod_{n=1}^\infty (1-(\frac{z}{n\pi})^2)$ converges and $\prod_{n=1}^\infty (1-(\frac{z}{n\pi}))$ doesn't (it seems to diverge to 0, as $\prod_{n=1}^\infty (1-(\frac{1}{n}))$ does), and $(1-(\frac{z}{\pi}))(1+(\frac{z}{\pi}))(1-(\frac{z}{2\pi}))(1+(\frac{z}{2\pi}))\dots$ converges conditionally, we see that we must use in the proof that $\sum_{n=1}^{\infty} a_n$ converges conditionally. Perhaps we can do this by showing if $a_n$ fluctuates between positive and negative, so does $\log(1+a_n)$, and so the infinite product converges conditionally.

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Edited to add:

As reminded by a comment, $a_n$ is real. Perhaps it being not complex is also a hint.

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Inspired by comments, I get proof 1. I put it as a solution.

My question for that proof: can we 'change order of summation' in equation (1) in my solution (considering we can't always do so for non absolutely convergent series)? Namely, is it correct that $$\sum_{n=N}^{N+p} ({a_n} -\frac{{a_n}^2}{2}+\frac{{a_n}^3}{3}+\dots)=\sum_{n=N}^{N+p}{a_n}-\sum_{n=N}^{N+p}\frac{{a_n}^2}{2}+\sum_{n=N}^{N+p}\frac{{a_n}^3}{3}+\dots,$$ and what if it's not 'N+p' but $\infty$. I guess it's a bit problematic because even if it's 'N+p', that 'long string' is still an infinite series, whose (absolute, conditional or non) convergence is not proven yet.

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I give another proof 2 using L'Hospital principle (in a way different from the hint of an answer) and post it as an answer.

My question is, this proof doesn't use that $\sum {a_n}^2$ converges, so according to the description of the problem, we can't get convergence of the infinite product without using convergence of $\sum {a_n}^2$. What's going wrong there?

I see one problem is since we have 'for all p, ...<$\epsilon$', then $\epsilon$ should be independent of p, but in my proof it is not!

More specifically, $\epsilon$ there depends on x only, not p, (so p depends not on $\epsilon$) so however small $\epsilon$ is, we can choose p sufficiently large, e.g. 1/$\epsilon$-1, so that $\epsilon'+\epsilon|M|(p+1)$ can't be arbitrarily small, but e.g. $>|M|$.

Answer 1

Hint: By L'Hopital's Rule $\frac {x-\log (1+x)} {x^{2}} \to \frac 1 2$ as $x \to 0$. Hence there exist positive constants $c_1,c_2$ and $\delta >0$ such that $c_1x^{2} \leq (x-log (1+x)) \leq c_2x^{2}$ for $-\delta <x <\delta$. The result follows easily from this and I will let you write out the details.

Answer 2

Proof 1.

I get a (thought not yet strict) proof.

$\sum_{n=N}^{N+p}\ \log(1+a_n) =\sum_{n=N}^{N+p} ({a_n} -\frac{{a_n}^2}{2}+\frac{{a_n}^3}{3}+\dots),$ (1)

where p is any natural number. (Though we have not proved the absolute convergence of the series $\sum_{n=N}^{\infty}\ \log(1+a_n)$, not to say the double series $\sum_{n=N}^{\infty}({a_n}-\frac{{a_n}^2}{2}+\frac{{a_n}^3}{3}+\dots)$), assume that we can change order of summation, so we get (1) equals

$$ \sum_{n=N}^{N+p}{a_n}-\sum_{n=N}^{N+p}\frac{{a_n}^2}{2}+\sum_{n=N}^{N+p}\frac{{a_n}^3}{3}+\dots.$$

For both $\sum{a_n}$ and $\sum{a_n}^2$ converges, so for all $\lambda, \eta >0$, there exists $N$ such that $\sum_{n=N}^{N+p}{a_n}\leq\lambda,\ \sum_{n=N}^{N+p}{a_n}^2\leq\eta$.

Also we have $\sum_{n=N}^{N+p}{{a_n}^{2k}}\leq \{\sum_{n=N}^{N+p}{{a_n}^2\}^k}\leq \eta^k$. Also, ${a_n}$ is bounded, that is, there exists $M$ such that $|{a_n}|\leq M$, and so $|\sum_{n=N}^{N+p}{{a_n}^{2k+1}}| \leq \sum_{n=N}^{N+p}{M{a_n}^{2k}}\leq {M\eta}$. So we have the absolute value of (1) less than or equal to

$\lambda+\frac{{\eta}}{2}+\frac{M{\eta}}{3}+\frac{{\eta}^2}{4}+\frac{M{\eta}^2}{5}\dots \leq \lambda+\frac{(1+M){\eta}}{2}+\frac{(1+M){\eta}^2}{4}\dots \leq \lambda +\frac{(1+M){\eta}}{2(1-\eta)},$ (2)

if we choose $\eta$ to be less than unity, or 1/2, for convenience.

For all $\epsilon$, we can choose $\lambda <\frac{\epsilon}{2}$, $\eta <\frac{\epsilon}{2(1+M)}$, such that (2) is less than $\epsilon$, and so

$\forall \epsilon, p, \exists N$, s.t. $|\sum_{n=N}^{N+p}\ \log(1+a_n)|<\epsilon$, that is, $\sum_{n=1}^{\infty}\ \log(1+a_n)$ converges, and so does $\prod_{n=1}^{\infty}\ (1+a_n)$.

Basically what I think here is to use the limits, approximately $\lambda$ and $\eta$, of $\sum{a_n}$ and $\sum{a_n}^2$, to represent the expanded double series (or a 1-dim 'long' series; summed up differently, by $a_n$'s powers) of $\sum \log(1+a_n)$.

And the key here is as $\sum a_n$ converges (thought not absolutely) (and so it's bounded), and $\sum {a_n}^2$ does too, the sum of the tail of the two series tends to be small, and so the tail in (2) tends to be small as well. (Notice if $a_n$ just tends to 0 and is bounded but $\sum a_n$ is not convergent, e.g. $\sum$1/n, we can't have the first item in (2) being arbitrarily small.

This proof may illustrate the relation between $\sum{a_n}^k$ and $\prod(1+a_n)$, e.g. it's easy to see the latter contains $\sum{a_n}$, for higher order items, the thing goes more complicated, namely we have more items (products of $a_i, a_j,\dots$, some with different indexes) than those in $\sum \log(1+a_n)$. It seems in taking log of an infinite product, we drop not only '1' but the just mentioned items.

Answer 3

Proof 2.

This proof is inspired by an answer, and uses 'parallel' relation between log(1+x) and x for small x.

Using L'Hospital principle, we have $\lim_{x\rightarrow 0} \frac{log(1+x)}{x}=1$ (notice here we 'drop' the 2nd and higher order items of log(1+x)), so for all $\epsilon>0$, given x sufficiently near 0, we have $x-\epsilon|x|<log(1+x)<x+\epsilon|x|$.

For $\sum a_n$ converges, given n sufficiently large, $a_n$ will be sufficiently near 0, and $\exists M$, $|a_n|<M$, and so $a_n-\epsilon|M|<log(1+a_n)<a_n+\epsilon|M|$. Therefore given N sufficiently large (so that n is), $\forall p, (\sum_{n=N}^{N+p} a_n)-\epsilon|M|(p+1)< \sum_{n=N}^{N+p}\ \log(1+a_n) < (\sum_{n=N}^{N+p} a_n)+\epsilon|M|(p+1)$.

For $\sum a_n$ converges, $\forall \epsilon'>0$, given N sufficiently large, $|\sum_{n=N}^{N+p} (a_n)|<\epsilon'$, and so $|\sum_{n=N}^{N+p}\ \log(1+a_n)| < \epsilon'+\epsilon|M|(p+1)$, which can be arbitrarily small by choosing $\epsilon, \epsilon'$ sufficiently small. So we have $\sum_{n=1}^{\infty}\ \log(1+a_n)$ converges, and therefore $\prod_{n=1}^{\infty}\ (1+a_n)$ converges (for 'exp(lim ...) = lim exp(...)').

But my question is, this proof doesn't use that $\sum {a_n}^2$ converges, so according to the description of the problem, we can't get convergence of the infinite product without using convergence of $\sum {a_n}^2$. What's going wrong here?

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The answer that I refer to uses $\frac {x-\log (1+x)} {x^{2}} \to \frac 1 2$ as $x \to 0$, which makes more sense, though I guess if we do so we would still 'drop out' higher order items, and have similar issues stated above?