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I'm just curious if this would work. The first products I looked at are trivial. I just want to know if more can be said about this one:

$$f(x)=\prod_{n=1}^\infty\dfrac{1}{1+\large\frac{(-1)^n}{nx}}$$

Does this now converge?

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user54358
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    Please try it yourself. What do you thing SE is? A place where you can freely give other people questions that you have not thought about? Please dont. – CBenni Jan 14 '13 at 16:35
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    It's a valuable learning experience. I'm not afraid to ask stupid questions. – user54358 Jan 14 '13 at 16:49
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    Its not about bad questions. Please keep in mind other people take their own time to answer this. Maybe put some thought into it yourself. – CBenni Jan 14 '13 at 17:16
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    The most important thing is to never stop asking questions. I don't understand why you sir are telling me to do the exact opposite! – user54358 Jan 14 '13 at 17:22
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    I think what CBenni is saying is that you should tell us what you tried. Do you know any criterion for a product to converge? Have you tried to apply it? What difficulties did you encounter? Etc. – jathd Jan 14 '13 at 18:13
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    Your question is not "stupid", but your attitude is. The purpose of MSE is to provide guide to users who already tried something to get the answer. Is good to never stop asking questions, but please, first ask to yourself. – Matemáticos Chibchas Jan 14 '13 at 18:26
  • Checked the "faqs" and "about". And what you're saying about the purpose of MSE is false. I think the question is answerable and I am curious about any explanation. Enough said. Anyway, I tried looking for a bound $\prod(1+a_n)$ because then convergence of $\sum a_n$ would imply convergence. – user54358 Jan 14 '13 at 18:54
  • @user54358 OK, now please share with us your thoughts about the convergence of the series $\sum a_n$, feel free to edit your original post to include this. – Matemáticos Chibchas Jan 15 '13 at 01:44

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So I just looked at convergence of the reciprocal.

$g(x)=\prod_{n=1}^\infty\left(1+\frac{(-1)^n}{nx}\right)$

Convergence of $g$ is equivalent to convergence of the sum

$\sum_{n=1}^\infty\frac{(-1)^n}{nx} = -\frac{1}{x}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n} = -\frac{ln2}{x}$

The last sum is the alternating harmonic series. Ok I don't know if this is correct. Because the sum is not defined for $x=0$ but the original product is undefined for $x=\frac{1}{n}$ where $n$ is odd.

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